User Log-Machine Name

2002-11-05 Thread PK_Deepa/VGIL
Hello

A user log is maintained for a table.I want to log all updations made to
this table.
I have written a database trigger (After Update) to log the changes to
another  table.My requirement is to find the machine name from which the
updation has occurred and log the machine name also .
How to get the currently working session ID so that machine name can be
retrieved fromV$session view or any other method to retrive the machine
name uniquely.

Is it that one Oracle user can have multiple active sessions on different
machines? Eg.
User Name Status Module  Session   Machine
XXACTIVESQL*Plus   10 AA
XXACTIVESQL*Plus   19 BB

Regards,
Deepa


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RE: Unkillable Background process SHUTDOWN ABORT LEAVES UNKILLAB

2002-11-05 Thread Sinardy Xing
Hi Stephen,



Sun Clustering basically quite simple (I think) :) 
an Oracle user dedicated to this software use to monitor (do query for certain system 
tables) the existence of the db
if anything wrong than this so call fault monitoring will do shutdown abort and 
release the mount points for 
second node to do startup pfile=

the problem is shutdown abort successful but the mounted points cannot be unmount 
because of those 
UNKILLED background processes.

I think sun clustering shuold add this 
if cannot release mount point then do STOP A follow by power off, right ?


Sinardy


-Original Message-
Sent: 01 November 2002 23:03
To: Multiple recipients of list ORACLE-L
UNKILLAB




 -Original Message-
 
 when I try to sync the session hung (init will sync 
 automatically therefore it hung too)
 I have to POWER OFF and POWER ON my SUN BOX to release the mount point
 My sun clustering fail to fail over because of this
 
 there is jargon in unix call zombie process (what is this ?) 
 What is tampered process, and when it happen ?
 

A zombie process is a process that was started by a parent process in such a
way that it is expected that the child process will exit and the parent will
acknowledge the exit; but the parent never did.  You used to see this all
the time on a Unix box in the past.  Now days, the init process usually
becomes the foster parent so you don't have zombies piling up as time goes
on.  There was never anything terribly wrong with zombies (as long as it
wasn't some ridiculous amount) since the only resource they tied up (at
least in theory) was a process ID number; and it made for a messy,
disconcerting ps output.

I know absolutely nothing about Sun clustering; but I am wondering if you
are having trouble with network file system reads and writes.  NFS problems
can cause hanging.
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RE: Relation Between Oracle Sessions

2002-11-05 Thread Stephane Faroult
Hussain,

   This is just from the top of my head and therefore requires careful checking, but I 
presume that if you are not using MTS then the 2nd generation session must be 
initiated by your shadow process (hope you are under Unix!). V$SESSION holds the OS 
process id of the program on the 'client side' (eq sqlplus), and V$PROCESS contains 
the details about the shadow processes (SADDR lets you link with V$SESSION). This is 
where I would have a look. If my theory is right, the 'middle generation' process id 
should be both in V$SESSION and V$PROCESS.

HTH

Stephane Faroult

- Original Message -
From: Hussain Ahmed Qadri [EMAIL PROTECTED]
To: Multiple recipients of list ORACLE-L
[EMAIL PROTECTED]
Sent: Mon, 04 Nov 2002 21:08:21

Hi all,
A question related to Developer but involves
forming sessions at the
backend, so would appreciate any help or ideas.

Whenever a form is opened, it opens an Oracle
session (can be seen from
v$session). Now when we call another form from this
opened form, (call form
property of forms), it will open another session.
If we consider the main
form as the parent form and the form called from
within it as its child
form, then, IS there a way to know, from some view
or else, that session of
this newly opened form is the child of which parent
session, their
relationship basically?

Thanks in advance

Hussain Ahmed Qadri
DBA SKMCHRC
Pakistan
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RE: Relation Between Oracle Sessions

2002-11-05 Thread Hussain Ahmed Qadri
Title: RE: Relation Between Oracle Sessions





Hi Justin,


We have our application in the form of a menu, with all the different modules written on top in the form of menus. Now with call-form property (which we use now in our application), only one form can be opened as the menu is disappeared from the top and is only made visible when that particular form is closed and this results in only one initial session. (Although by passing a parameter to call-form property of form, we can still tell Oracle to open this form in another session, other than the initial session). 

But when we use the Open-form property in forms, then every form that is opened from with in the application is opened in another session. Menu is there at the top and even if one form is open, you can use the menu and open another form or module.

So in my opinion that's not a bug but an Oracle property. And it's this newly opened session (if we use the open-form property) from with in the main application, whose relation to the session of the main application I wanted to know.

You might be absolutely right that Oracle doesn't keep track of the so-called parent-child relation ship of sessions, so I wanted to know if there was any indirect method, like from some views or something that can help me track that relationship. 

If there is something that is confusing, please let me know.
Thank you for your time.


Regards,


Hussain Ahmed Qadri
DBA
SKMCHRC


-Original Message-
From: Justin Cave [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, November 05, 2002 11:03 AM
To: Multiple recipients of list ORACLE-L
Subject: Re: Relation Between Oracle Sessions


At 10:08 PM 11/4/2002, you wrote:


Whenever a form is opened, it opens an Oracle session (can be seen from
v$session). Now when we call another form from this opened form, (call
form property of forms), it will open another session. If we consider the
main form as the parent form and the form called from within it as its
child form, then, IS there a way to know, from some view or else, that
session of this newly opened form is the child of which parent session,
their relationship basically?


I believe the simple answer here is no. If your application is opening
multiple sessions, each session is logically independent of the other--
Oracle has no concept of parent or child sessions. I would generally
take this sort of behavior to indicate a bug in your application--
generally one wants to reuse connections across multiple forms so that each
form doesn't have the overhead of opening a new connection to the database.



Justin Cave
Distributed Database Consulting


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ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE

2002-11-05 Thread Yechiel Adar
Hello All

Oracle 8.1.6.3.4 on NT.

I got ora-19502 - can not write archive log, and users could not logged on.
I can logon only as internal.
We had the same problem a few days ago.
The technical support people checked the raid 5 disks and did not find any
I/O errors.
There is a lot of free space on the drive.

Has anybody encountered this problem?
Do you know what can cause it?
What I need to do in order to prevent this from happening again?


Yechiel Adar
Mehish

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RE: User Log-Machine Name

2002-11-05 Thread Nirmal Kumar Muthu Kumaran
Hei...

Inside trigger block, using userenv('terminal') you can identify the machine
name.

One oracle user can have multiple active sessions. 

(Here in my team, 5 people working on the same oracle username with
different machines. 
The privileges are set for them at application level.)

HTH.

Nirmal.,


-Original Message-
[mailto:PK_Deepa/VGIL;vguard.satyam.net.in]
Sent: Tuesday, November 05, 2002 11:23 AM
To: Multiple recipients of list ORACLE-L


Hello

A user log is maintained for a table.I want to log all updations made to
this table.
I have written a database trigger (After Update) to log the changes to
another  table.My requirement is to find the machine name from which the
updation has occurred and log the machine name also .
How to get the currently working session ID so that machine name can be
retrieved fromV$session view or any other method to retrive the machine
name uniquely.

Is it that one Oracle user can have multiple active sessions on different
machines? Eg.
User Name Status Module  Session   Machine
XXACTIVESQL*Plus   10 AA
XXACTIVESQL*Plus   19 BB

Regards,
Deepa


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RE: User Log-Machine Name

2002-11-05 Thread Stephane Faroult
 audsid = SYS_CONTEXT('USERENV', 'SESSIONID')

You could have guessed it :-).

- Original Message -
From: [EMAIL PROTECTED]
To: Multiple recipients of list ORACLE-L
[EMAIL PROTECTED]
Sent: Tue, 05 Nov 2002 00:23:28

Hello

A user log is maintained for a table.I want to log
all updations made to
this table.
I have written a database trigger (After Update) to
log the changes to
another  table.My requirement is to find the
machine name from which the
updation has occurred and log the machine name also
.
How to get the currently working session ID so that
machine name can be
retrieved fromV$session view or any other method to
retrive the machine
name uniquely.

Is it that one Oracle user can have multiple active
sessions on different
machines? Eg.
User Name Status Module  Session  
Machine
XXACTIVESQL*Plus   10 AA
XXACTIVESQL*Plus   19 BB

Regards,
Deepa


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Regards,

Stephane Faroult
Oriole
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OCP Exams - What to study out of?

2002-11-05 Thread Denham Eva
Title: OCP Exams - What to study out of?





Hello Everyone,


1. I have now made the humungous decision to start studying and to write the OCP exams.
2. Do I study out of the Oracle Manuals? 
3. I do have the Sybex Study Guides, would studying these be all that is needed?
4. Or do the questions come out of the Course material ( Which I have not attended ).
5. I do have some of the Oreilly Insect Books - would these be any help?


Any views, opinions etc appreciated.
Regards
Denham Eva
Oracle DBA
UNIX is basically a simple operating system, but you have to be a genius to understand the simplicity.
Dennis Ritchie.






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suggestion w/c platforms to choose from...

2002-11-05 Thread grace
hi,
 can anyone suggestion w/c platform should i used to run oracle? wat are the
things to consider in choosing platform?

thanks

Best regards,
Grace Lim
MIS Department
Suy Sing Comm'l Corp.
T- (632)-2474134
F- (632)-2474160

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Read Only Login with Source Visibility

2002-11-05 Thread Bill Buchan
Hi all,

I am trying to create a read-only login, RO_USER which can do the following:

1. See all tables, views, constraints etc. in one  *specific* other schema, 
APP.

2. See all the source for PL/SQL objects in APP.

The first bit is easy: GRANT SELECT on tables, views to RO_USER.

Not sure about the second bit.  I have tried granting CREATE ANY PROCEDURE 
and CREATE ANY TRIGGER to RO_USER and using a DDL trigger to prevent these 
privileges actually being used to create procedures/triggers.  This works 
but does not restrict the source visibility to APP.  I have other schemas 
where I do not want RO_USER to see the source.

My other concern is granting SELECT on sequences as this means that they 
can select nextval from them and hence increment the numbers (not quite 
read-only!)

Any suggestions for fixing these problems would be much appreciated!

Thanks
- Bill.

PS. This is on 8i.


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Re: suggestion w/c platforms to choose from...

2002-11-05 Thread Yechiel Adar
As a rule buy the biggest, meanest, fault tolerance, with gigabytes of
memory and terabytes of disk storage that you can buy.

If you will provide more data about:

1) The size of the database
2) How many users
3) How critical is the system
4) The use of the system - data warehouse, OLTP etc

then you will probably get a more specified answer.

Yechiel Adar
Mehish
- Original Message -
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 2:13 PM


 hi,
  can anyone suggestion w/c platform should i used to run oracle? wat are
the
 things to consider in choosing platform?

 thanks

 Best regards,
 Grace Lim
 MIS Department
 Suy Sing Comm'l Corp.
 T- (632)-2474134
 F- (632)-2474160

 --
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 --
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RE: suggestion w/c platforms to choose from...

2002-11-05 Thread Mark Leith
How about considering what OS skills your systems admins already have? Do
they have experience with HP/UX, or Solaris, or NT/2000 type systems etc.?

How big is the database? How many users will it have? What will the database
be used for?

How much money do you have to spend?

Mark

-Original Message-
Sent: 05 November 2002 12:14
To: Multiple recipients of list ORACLE-L


hi,
 can anyone suggestion w/c platform should i used to run oracle? wat are the
things to consider in choosing platform?

thanks

Best regards,
Grace Lim
MIS Department
Suy Sing Comm'l Corp.
T- (632)-2474134
F- (632)-2474160

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RE: OCP Exams - What to study out of?

2002-11-05 Thread Paula_Stankus
Title: OCP Exams - What to study out of?



I only 
studied the Coriolius books and done the self-exams (which are harder then the 
real one) and passed with 85+% each time. I think the course materials 
were not as concise as the Cor. books. I believe you can get some of those 
books still on Amazon. That publisher is out of business. Someone 
recently posted which publisher is taking over for them. Good 
luck.

  -Original Message-From: Denham Eva 
  [mailto:[EMAIL PROTECTED]]Sent: Tuesday, November 05, 2002 7:09 
  AMTo: Multiple recipients of list ORACLE-LSubject: OCP 
  Exams - What to study out of?
  Hello Everyone, 
  1. I have now made the humungous decision to start 
  studying and to write the OCP exams. 2. Do 
  I study out of the Oracle Manuals? 3. I do 
  have the Sybex Study Guides, would studying these be all that is 
  needed? 4. Or do the questions come out of 
  the Course material ( Which I have not attended ). 5. I do have some of the Oreilly "Insect" Books - would these be any 
  help? 
  Any views, opinions etc appreciated. 
  Regards Denham 
  Eva Oracle DBA 
  "UNIX is basically a simple 
  operating system, but you have to be a genius to understand the 
  simplicity." Dennis 
  Ritchie. 
  
  
  DISCLAIMER 
  
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  mistransmission. If you receive this message in error, please immediately 
  delete it and all copies of it from your system, destroy any hard copies of it 
  and notify the sender. You must not, directly or indirectly, use, disclose, 
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ORA-02050

2002-11-05 Thread Ray Stell

ORA-02050: transaction 8.82.26033 rolled back, some remote DBs may be in-doubt

The transaction seems to have completely rolled back.  My 
question is, is there any way to relate transaction id 8.82.26033
to an application or table row or something at a higher layer?

Thanks.
===
Ray Stell   [EMAIL PROTECTED] (540) 231-4109 KE4TJC28^D
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Re: Read Only Login with Source Visibility

2002-11-05 Thread Tim Gorman
How about redefining copies of the relevant data dictionary views for
RO_USER to suit your purposes?

- Original Message -
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 5:48 AM


 Hi all,

 I am trying to create a read-only login, RO_USER which can do the
following:

 1. See all tables, views, constraints etc. in one  *specific* other
schema,
 APP.

 2. See all the source for PL/SQL objects in APP.

 The first bit is easy: GRANT SELECT on tables, views to RO_USER.

 Not sure about the second bit.  I have tried granting CREATE ANY PROCEDURE
 and CREATE ANY TRIGGER to RO_USER and using a DDL trigger to prevent these
 privileges actually being used to create procedures/triggers.  This works
 but does not restrict the source visibility to APP.  I have other schemas
 where I do not want RO_USER to see the source.

 My other concern is granting SELECT on sequences as this means that they
 can select nextval from them and hence increment the numbers (not quite
 read-only!)

 Any suggestions for fixing these problems would be much appreciated!

 Thanks
 - Bill.

 PS. This is on 8i.


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RE: OCP Exams - What to study out of?

2002-11-05 Thread John . Hallas
Denham,
I have just passed my 8i upgrade exam and added some notes to a page about
getting OCP certification.
It can be accessed on http://www.hcresources.co.uk/ocp.htm
http://www.hcresources.co.uk/ocp.htm  
 
Hope you find it interesting
 
John

-Original Message-
Sent: 05 November 2002 12:09
To: Multiple recipients of list ORACLE-L



Hello Everyone, 

1. I have now made the humungous decision to start studying and to write the
OCP exams. 
2. Do I study out of the Oracle Manuals? 
3. I do have the Sybex Study Guides, would studying these be all that is
needed? 
4. Or do the questions come out of the Course material ( Which I have not
attended ). 
5. I do have some of the Oreilly Insect Books - would these be any help? 

Any views, opinions etc appreciated. 
Regards 
Denham Eva 
Oracle DBA 
UNIX is basically a simple operating system, but you have to be a genius to
understand the simplicity. 
Dennis Ritchie. 



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RTIF Interface

2002-11-05 Thread KENNETH JANUSZ



What is an RTIF interface? I looked on Acronym Finder 
and it didn't find it in it's database.

Thanks,
Ken Janusz, CPIM


RE: ORA-02050

2002-11-05 Thread John . Hallas
Ray,
I think you need to check DBA_2PC_PENDING on both the local and remote
database to see if anything is still in there.
That gives the osuser, terminal and host name so you may be able to get
something from that.
Of course if there is nothing in that table then the transaction has been
rolled back already 

John

-Original Message-
Sent: 05 November 2002 13:43
To: Multiple recipients of list ORACLE-L



ORA-02050: transaction 8.82.26033 rolled back, some remote DBs may be
in-doubt

The transaction seems to have completely rolled back.  My 
question is, is there any way to relate transaction id 8.82.26033
to an application or table row or something at a higher layer?

Thanks.
===
Ray Stell   [EMAIL PROTECTED] (540) 231-4109 KE4TJC28^D
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Re: Read Only Login with Source Visibility

2002-11-05 Thread Dale
Hi Bill

 I am trying to create a read-only login, RO_USER which can do the
following:
 1. See all tables, views, constraints etc. in one  *specific* other
schema,
 2. See all the source for PL/SQL objects in APP.

Is what you really want is a way for users to be able to see the DDL and
PL/SQL without actually being able to run or modify it? If so, as a
suggestion, why don't you try using the DDL to HTML documentation generator
in the freeware DBATool. You could easily cut a browsable tree with all (or
part) of a schemas DDL. It also does nice things like cross reference
related entities - for example for a table you can click on links to the
associated indexes, foreign keys, grants, synonyms etc.

Password protection and security can easily be implemented by moving the
HTML tree onto a password protected drive.

An example tree can be found at the link below

http://www.databee.com/dt_htmltree/db=TESTDB1+sc=DT_TEST.htm

The DBATool is freeware http://www.DataBee.com

Regards
Dale Edgar
Net 2000 Ltd.
[EMAIL PROTECTED]

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RE: MTS process HIGH CPU

2002-11-05 Thread Gogala, Mladen
That process has an oracle session associated with it.
Before killing it, it would be nice to see what that 
session is doing. You could do it by joining v$process and
v$session. The SPID column in v$process is System PID and
ADDR column corresponds to the PADDR in V$SESSION table.

 -Original Message-
 From: Seema Singh [mailto:oracledbam;hotmail.com]
 Sent: Monday, November 04, 2002 2:09 PM
 To: Multiple recipients of list ORACLE-L
 Subject: MTS process HIGH CPU
 
 
 Hi
 I am using MTS in 8.1.6.I am wondering one of MTS process is 
 locked 2 table 
 since morning and this process is taking more CPU.Can I kill 
 that process 
 like ora_s000_prod by using alter system and then unix kill command?
 Thx
 -Seema
 
 
 
 
 
 
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RE: Unkillable Background process SHUTDOWN ABORT LEAVES UNKILLAB

2002-11-05 Thread Stephen Lee

I once used a product like this called First Watch, but I called it
automatic failover rather than clustering.  The product provided a redundant
heart beat monitoring; but the user had to set up the configuration files
and write and maintain any scripts that controlled the failover.  What we
discovered was that the Sun OS and hardware were so reliable that the most
likely point of failure was in the human management of the configuration
files and scripts.  That is, the hardware was more reliable than people.  So
we decided to disable the automatic failover and use manual failover, where
we would manually unplug the cables to the storage arrays from one box, plug
them into the failover box, then boot the failover box by hand.

We never had to use it.  What continues to amaze me is how often people
implement these complex and difficult to manage redundancy setups (sometimes
referred to NOT as a cluster, but a cluster f**k) when, if they would just
choose a reliable platform, they would be better off.


 -Original Message-
 if anything wrong than this so call fault monitoring will do 
 shutdown abort and release the mount points for 
 second node to do startup pfile=
 
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Clone Production Server to Stand by Server on 8.1.7 on Win 2k

2002-11-05 Thread Arif Khan (GWL)
Hello

We currently have two Identical servers (identical in terms of both Hardware
and SW).
We run Oracle 7.3.4.0 (Workgroup server) on Win NT 4 on both the servers. We
call them as Production Server and Stand by server. The Stand by server is
passive in nature (i.e. does not do anything). Every night a batch process
shuts down Oracle instance on both the machines and copies over all the
files (Data, log, ctl etc) from Production Server to the Stand by server
(Drive to drive, directory to directory...)

In case the Production Server fails, we simply switch over the users (with a
different alias to the stand by server) and they are back in business.

Now, we are thinking of migrating to 8.1.7, however while trying to install
this version, one needs to specify a Global name which I believe has to be
unique on the network. So will the same process that I used to run (i.e.
copy all files over from production to stand by ) work??? I guess both my
servers will now have to have separate/unique Global Names. 

Is there any other approach that any of you can suggest??? 

TIA
Arif
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RE: OCP Exams - What to study out of?

2002-11-05 Thread Paula_Stankus
Title: RE: OCP Exams - What to study out of?





Guys, anyone know about an application called Cloverleaf? Basically, my organization is pursuing using Cloverleaf as the data integration sole application of choice. To me Unix/SQL*Loader and native tools for the various RDBMS usually are more straightforward and provide simplication and less proprietary coding and it is easier for an organization to find others who have those type of skills. Any ideas guys? 

-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, November 05, 2002 9:18 AM
To: Multiple recipients of list ORACLE-L
Subject: RE: OCP Exams - What to study out of?



Denham,
I have just passed my 8i upgrade exam and added some notes to a page about
getting OCP certification.
It can be accessed on http://www.hcresources.co.uk/ocp.htm
http://www.hcresources.co.uk/ocp.htm 

Hope you find it interesting

John


-Original Message-
Sent: 05 November 2002 12:09
To: Multiple recipients of list ORACLE-L




Hello Everyone, 


1. I have now made the humungous decision to start studying and to write the
OCP exams. 
2. Do I study out of the Oracle Manuals? 
3. I do have the Sybex Study Guides, would studying these be all that is
needed? 
4. Or do the questions come out of the Course material ( Which I have not
attended ). 
5. I do have some of the Oreilly Insect Books - would these be any help? 


Any views, opinions etc appreciated. 
Regards 
Denham Eva 
Oracle DBA 
UNIX is basically a simple operating system, but you have to be a genius to
understand the simplicity. 
Dennis Ritchie. 




 _ 


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confidential, proprietary or legally privileged information. No
confidentiality or privilege is waived or lost by any mistransmission. If
you receive this message in error, please immediately delete it and all
copies of it from your system, destroy any hard copies of it and notify the
sender. You must not, directly or indirectly, use, disclose, distribute,
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Re: ORA-02050

2002-11-05 Thread Ray Stell
On Tue, Nov 05, 2002 at 06:33:44AM -0800, [EMAIL PROTECTED] wrote:
 Ray,
 I think you need to check DBA_2PC_PENDING on both the local and remote
 database to see if anything is still in there.
 That gives the osuser, terminal and host name so you may be able to get
 something from that.
 Of course if there is nothing in that table then the transaction has been
 rolled back already 


Right, nothing there, so I was wondering if there was a way to chase
the transaction id to app, such as archivelog.   8.1.7.4 here.

Thanks.





 John
 
 -Original Message-
 Sent: 05 November 2002 13:43
 To: Multiple recipients of list ORACLE-L
 
 
 
 ORA-02050: transaction 8.82.26033 rolled back, some remote DBs may be
 in-doubt
 
 The transaction seems to have completely rolled back.  My 
 question is, is there any way to relate transaction id 8.82.26033
 to an application or table row or something at a higher layer?
 
 Thanks.
 ===
 Ray Stell   [EMAIL PROTECTED] (540) 231-4109 KE4TJC28^D
 -- 
 Please see the official ORACLE-L FAQ: http://www.orafaq.com
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Re: Clone Production Server to Stand by Server on 8.1.7 on Win 2k

2002-11-05 Thread Yechiel Adar
Hello

You did not write if you bring both databases on every night or the backup
is down until needed.
If it is down you have no problem.
If you bring both up you can change the global name :
Alter database rename global_name to backupdb;

BTW - I am not sure that 8.1.7 need global_name=true.
We are on 8.1.6 on NT and we do not use global names = true.

Yechiel Adar
Mehish
- Original Message -
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 4:53 PM


 Hello

 We currently have two Identical servers (identical in terms of both
Hardware
 and SW).
 We run Oracle 7.3.4.0 (Workgroup server) on Win NT 4 on both the servers.
We
 call them as Production Server and Stand by server. The Stand by server is
 passive in nature (i.e. does not do anything). Every night a batch process
 shuts down Oracle instance on both the machines and copies over all the
 files (Data, log, ctl etc) from Production Server to the Stand by server
 (Drive to drive, directory to directory...)

 In case the Production Server fails, we simply switch over the users (with
a
 different alias to the stand by server) and they are back in business.

 Now, we are thinking of migrating to 8.1.7, however while trying to
install
 this version, one needs to specify a Global name which I believe has to be
 unique on the network. So will the same process that I used to run (i.e.
 copy all files over from production to stand by ) work??? I guess both my
 servers will now have to have separate/unique Global Names.

 Is there any other approach that any of you can suggest???

 TIA
 Arif
 --
 Please see the official ORACLE-L FAQ: http://www.orafaq.com
 --
 Author: Arif Khan (GWL)
   INET: [EMAIL PROTECTED]

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 also send the HELP command for other information (like subscribing).

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RE: OCP Exams - What to study out of?

2002-11-05 Thread DENNIS WILLIAMS
Eva - Everything I've heard says that all the exam questions come out of the
Oracle University class Student Guides. To put it bluntly, your goal is to
pass the exam. You could study the Oracle manuals, but there is a lot of
material to cover and what strikes you as important might not be covered on
the exam, and vice-versa. Essentially you are getting two levels of
filtering. Somebody read the manuals and used that to prepare student
guides, then a group of Oracle instructors read the student guide and
prepared questions. I think that most of the exam guides were prepared from
the Oracle Student Guides and the author has at least taken the exam, so
they have a general idea of how the test is approached. Everyone has their
favorite exam preparation book and I consider this a worthwhile investment,
I bought Couchman -
http://search.barnesandnoble.com/booksearch/isbnInquiry.asp?userid=6WIANMIL0
H
http://search.barnesandnoble.com/booksearch/isbnInquiry.asp?userid=6WIANMIL
0Hsourceid=00400731866779927243bfdate=11%2D05%2D2002+10%3A16%3A17isbn=007
2133414
sourceid=00400731866779927243bfdate=11%2D05%2D2002+10%3A16%3A17isbn=00721
33414
   Two issues: Which exam track are you planning to take? 9i? 8i? 
   The advice I received is don't take the SQL exam as your first exam.
Take the DBA exam first. The SQL exam is reported to be quite hard because
it covers all the odd SQL functions and can really catch you unawares. As a
practicing DBA you should find the DBA exam pretty easy. I didn't personally
take that path for my own reasons, but I still consider it good advice.



Dennis Williams 
DBA, 40%OCP 
Lifetouch, Inc. 
[EMAIL PROTECTED] 

-Original Message-
Sent: Tuesday, November 05, 2002 6:09 AM
To: Multiple recipients of list ORACLE-L



Hello Everyone, 

1. I have now made the humungous decision to start studying and to write the
OCP exams. 
2. Do I study out of the Oracle Manuals? 
3. I do have the Sybex Study Guides, would studying these be all that is
needed? 
4. Or do the questions come out of the Course material ( Which I have not
attended ). 
5. I do have some of the Oreilly Insect Books - would these be any help? 

Any views, opinions etc appreciated. 
Regards 
Denham Eva 
Oracle DBA 
UNIX is basically a simple operating system, but you have to be a genius to
understand the simplicity. 
Dennis Ritchie. 



  _  

DISCLAIMER 



This message is for the named person's use only. It may contain
confidential, proprietary or legally privileged information. No
confidentiality or privilege is waived or lost by any mistransmission. If
you receive this message in error, please immediately delete it and all
copies of it from your system, destroy any hard copies of it and notify the
sender. You must not, directly or indirectly, use, disclose, distribute,
print, or copy any part of this message if you are not the intended
recipient. TFMC, its holding company, and any of its subsidiaries each
reserve the right to monitor and manage all e-mail communications through
its networks. 

Any views expressed in this message are those of the individual sender,
except where the message states otherwise and the sender is authorized to
state them to be views of any such entity. 

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Enterprise Manager Console - Export problem

2002-11-05 Thread Arif Khan (GWL)
Hello

I run Oracle 8.1.6 on Win 2k (Both standard ed.). I am trying to run the
export wizard through the Enterprise Manager Console by doing the following:
Under Database folder, I right click the Global Name (say ORCL) - Data
Management - Export ...
and it comes back with the error Either Preferred Credentials are not set or
the username and/or password are invalid for this database and node. You
must set the Preferred Credentials for the database and node in the Oracle
Enterprise Manager Console OK

I have logged in to EM-Console with SYSMAN user. Under System - Preferences
- Preferred Credential, for ORCL, I have tried using SYSMAN, SYSTEM and SYS
with Role as NORMAL, SYSDBA  SYSOPER.

I have no credentials setup  for the Node (say ORATEST).

I am pretty new to Oracle 8i and hence may not be taking the right steps
required.

TIA
Arif
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Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE

2002-11-05 Thread Jared . Still
Yechiel,

What appears in the archive log?

There is very likely to be some OS errors included.

Jared






Yechiel Adar [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 12:58 AM
 Please respond to ORACLE-L

 
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
cc: 
Subject:ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE


Hello All

Oracle 8.1.6.3.4 on NT.

I got ora-19502 - can not write archive log, and users could not logged 
on.
I can logon only as internal.
We had the same problem a few days ago.
The technical support people checked the raid 5 disks and did not find any
I/O errors.
There is a lot of free space on the drive.

Has anybody encountered this problem?
Do you know what can cause it?
What I need to do in order to prevent this from happening again?


Yechiel Adar
Mehish

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Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE

2002-11-05 Thread Yechiel Adar
Here is parts of the alert log:

  Current log# 6 seq# 4153 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1356.ORA
Sun Nov 03 11:03:34 2002
ARC0: Beginning to archive log# 5 seq# 4152
ARC0: Completed archiving log# 5 seq# 4152
Sun Nov 03 11:10:33 2002
Thread 1 advanced to log sequence 4154
  Current log# 7 seq# 4154 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1357.ORA
Sun Nov 03 11:10:33 2002
ARC2: Beginning to archive log# 6 seq# 4153
ARC2: Completed archiving log# 6 seq# 4153
Sun Nov 03 11:20:50 2002
Thread 1 advanced to log sequence 4155
  Current log# 8 seq# 4155 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1358.ORA
Sun Nov 03 11:20:50 2002
ARC1: Beginning to archive log# 7 seq# 4154
ARC1: I/O error 19502 archiving log 7 to
'E:\ORACLE\ORADATA\MUSK135\DATABASE\ARCHIVE\M135T001S04154.ARC'
ARC1: Archiving not possible: error count exceeded
ARC1: Failed to archive log# 7 seq# 4154
ARCH: Archival stopped, error occurred. Will continue retrying
ARCH:
 ORA-16038: log 7 sequence# 4154 cannot be archived
ORA-19502: write error on file , blockno  (blocksize=)
ORA-00312: online log 7 thread 1:
'D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1357.ORA'
Sun Nov 03 11:28:06 2002
Thread 1 advanced to log sequence 4156
  Current log# 3 seq# 4156 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1353.ORA
Sun Nov 03 11:28:06 2002
ARC2: Beginning to archive log# 7 seq# 4154
ARC2: Archiving not possible: No primary destinations
ARC2: Failed to archive log# 7 seq# 4154
ARCH: Archival stopped, error occurred. Will continue retrying
ARCH:
 ORA-16014: log 7 sequence# 4154 not archived, no available destinations
ORA-00312: online log 7 thread 1:
'D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1357.ORA'
Sun Nov 03 11:38:54 2002
Thread 1 advanced to log sequence 4157
  Current log# 9 seq# 4157 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOG9.ORA
Sun Nov 03 11:38:55 2002
ARC0: Beginning to archive log# 7 seq# 4154
ARC0: Archiving not possible: No primary destinations
ARC0: Failed to archive log# 7 seq# 4154
Sun Nov 03 11:39:07 2002
Thread 1 advanced to log sequence 4158
  Current log# 10 seq# 4158 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOG10.ORA
Sun Nov 03 11:39:07 2002
ARC1: Beginning to archive log# 7 seq# 4154
ARC1: Archiving not possible: No primary destinations
ARC1: Failed to archive log# 7 seq# 4154

Yechiel Adar
Mehish
- Original Message -
To: ORACLE-L [EMAIL PROTECTED]
Cc: Yechiel Adar [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 7:31 PM


 Yechiel,

 What about those alert log entries?

 Jared






 Yechiel Adar [EMAIL PROTECTED]
  11/05/2002 10:38 AM


 To: ORACLE-L [EMAIL PROTECTED],
[EMAIL PROTECTED],
 [EMAIL PROTECTED]
 cc:
 Subject:Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE
SPACE


 Hello Charlie and Jared

 Thanks for the reply.

 As I mentioned in my mail there is a lot of free space on the disk and the
 database is with auto archive.
 Our problem is that the ARCH process gives an error message (ora 19502)
 and
 stop. There are messages on the console about error trying to write the
 archive log, like when the disk is full. Then the redo log will fill up
 and
 the instance will stop.

 After I connect as internal and do:
 archive log stop;
 archive log all;
 archive log start;

 everything is back to normal.

 Yechiel Adar
 Mehish
 - Original Message -
 From: [EMAIL PROTECTED]
 To: [EMAIL PROTECTED]
 Sent: Tuesday, November 05, 2002 4:48 PM
 Subject: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE


  My first guess is that the instance has archiving enable, but the
 ARCHIVER
  is/was not started.
  The DB will run OK until all the redo logfiles have been filled.
  At this point the instance will simply cease to process any additional
  transactions
  until it has space to which to write additional redo logfile
 information.
 
  There should be an entry in the alert_SID.log about what is happening 
  why.
 
  - Forwarded by Charlie Mengler/THD on 11/05/2002 06:45 AM -
 
Yechiel Adar
[EMAIL PROTECTED]To:   Multiple
 recipients of list ORACLE-L [EMAIL PROTECTED]
.il cc:
Sent by: Subject:  ORA-19502 ON
 RAID
 5 DISKS WITH ENOUGH FREE SPACE
[EMAIL PROTECTED]

 
 
11/05/2002 12:58
AM
Please respond to
ORACLE-L
 
 
 
 
 
 
  Hello All
 
  Oracle 8.1.6.3.4 on NT.
 
  I got ora-19502 - can not write archive log, and users could not logged
 on.
  I can logon only as internal.
  We had the same problem a few days ago.
  The technical support people checked the raid 5 disks and did not find
 any
  I/O errors.
  There is a lot of free space on the drive.
 
  Has anybody encountered this problem?
  Do you know what can cause it?
  What I need to do in order to prevent this from happening 

SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
id  number  not null 
constraint pk_treenode primary key,
parentidnumber  not null,
nodeorder   number  not null,
description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2
-

Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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Re: suggestion w/c platforms to choose from...

2002-11-05 Thread Jared . Still
How about a 400 CPU nCube running Oracle 9i RAC?

If they still support nCube that is. ( Larry owns nCube )


Seriously though, does your app maintain HR data for 500 people,
or is it a 20TB Data Warehouse?

These apps may require different HW configurations.

Jared






grace [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 04:13 AM
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cc: 
Subject:suggestion w/c platforms to choose from...


hi,
 can anyone suggestion w/c platform should i used to run oracle? wat are 
the
things to consider in choosing platform?

thanks

Best regards,
Grace Lim
MIS Department
Suy Sing Comm'l Corp.
T- (632)-2474134
F- (632)-2474160

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Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE

2002-11-05 Thread Yechiel Adar
Hello Charlie and Jared

Thanks for the reply.

As I mentioned in my mail there is a lot of free space on the disk and the
database is with auto archive.
Our problem is that the ARCH process gives an error message (ora 19502) and
stop. There are messages on the console about error trying to write the
archive log, like when the disk is full. Then the redo log will fill up and
the instance will stop.

After I connect as internal and do:
archive log stop;
archive log all;
archive log start;

everything is back to normal.

Yechiel Adar
Mehish
- Original Message -
To: [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 4:48 PM


 My first guess is that the instance has archiving enable, but the ARCHIVER
 is/was not started.
 The DB will run OK until all the redo logfiles have been filled.
 At this point the instance will simply cease to process any additional
 transactions
 until it has space to which to write additional redo logfile information.

 There should be an entry in the alert_SID.log about what is happening 
 why.

 - Forwarded by Charlie Mengler/THD on 11/05/2002 06:45 AM -

   Yechiel Adar
   [EMAIL PROTECTED]To:   Multiple
recipients of list ORACLE-L [EMAIL PROTECTED]
   .il cc:
   Sent by: Subject:  ORA-19502 ON RAID
5 DISKS WITH ENOUGH FREE SPACE
   [EMAIL PROTECTED]



   11/05/2002 12:58
   AM
   Please respond to
   ORACLE-L






 Hello All

 Oracle 8.1.6.3.4 on NT.

 I got ora-19502 - can not write archive log, and users could not logged
on.
 I can logon only as internal.
 We had the same problem a few days ago.
 The technical support people checked the raid 5 disks and did not find any
 I/O errors.
 There is a lot of free space on the drive.

 Has anybody encountered this problem?
 Do you know what can cause it?
 What I need to do in order to prevent this from happening again?


 Yechiel Adar
 Mehish

 --
 Please see the official ORACLE-L FAQ: http://www.orafaq.com
 --
 Author: Yechiel Adar
   INET: [EMAIL PROTECTED]

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 also send the HELP command for other information (like subscribing).





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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Jamadagni, Rajendra
Title: RE: SQL Brain Teaser Challenge





SELECT LEVEL, treenode.*
 FROM treenode 
START WITH parentid=0 
CONNECT BY PRIOR ID = parentid
ORDER SIBLINGS BY PARENTid , nodeorder


Raj
__
Rajendra Jamadagni  MIS, ESPN Inc.
Rajendra dot Jamadagni at ESPN dot com
Any opinion expressed here is personal and doesn't reflect that of ESPN Inc. 
QOTD: Any clod can have facts, but having an opinion is an art!



-Original Message-
From: Orr, Steve [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, November 05, 2002 12:24 PM
To: Multiple recipients of list ORACLE-L
Subject: SQL Brain Teaser Challenge



Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:


create table treenode (
 id  number  not null 
   constraint pk_treenode primary key,
 parentid number  not null,
 nodeorder number  not null,
 description varchar2(20)  null);


insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
 ID PARENTID NODEORDER DESCRIPTION
-- -- -- 
 1 0 0 top folder
 9 1 0 1st subfolder
 7 1 2 3rd subfolder
 8 7 1 folder 3 item 2
 5 7 0 folder 3 item 1
 2 1 1 2nd subfolder
 6 2 3 folder 2 item 3
 3 2 2 folder 2 item 2
 4 2 1 folder 2 item 1
-
Desired SQL statement results:
 ID PARENTID NODEORDER DESCRIPTION
-- -- -- 
 1 0 0 top folder
 9 1 0 1st subfolder
 2 1 1 2nd subfolder
 4 2 1 folder 2 item 1
 3 2 2 folder 2 item 2
 6 2 3 folder 2 item 3
 7 1 2 3rd subfolder
 5 7 0 folder 3 item 1
 8 7 1 folder 3 item 2
-


Kudos to anyone who can figure out how to do this via SQL.



Steve Orr
Bozeman, Montana
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This e-mail 
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Re: Clone Production Server to Stand by Server on 8.1.7 on Win 2k

2002-11-05 Thread Jared . Still
First off, your process is doing a lot of unnecessary work.

Standby databases are available in 7.3.4.  I believe that
would be somewhat simpler than your current procedure.

I haven't tried it though, so I could be wrong.

Even with your current procedure, you don't need to copy
all of the files, most of the time anyway.

Build your standby database, put it in recovery mode, and 
just keep applying archive log files to it.

You may need to rebuild when a datafile is added to a 
tablespace, not sure.

In 8i, why not just use standby database?

Jared





Arif Khan (GWL) [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 06:53 AM
 Please respond to ORACLE-L

 
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
cc: 
Subject:Clone Production Server to Stand by Server on 8.1.7 on Win 2k


Hello

We currently have two Identical servers (identical in terms of both 
Hardware
and SW).
We run Oracle 7.3.4.0 (Workgroup server) on Win NT 4 on both the servers. 
We
call them as Production Server and Stand by server. The Stand by server is
passive in nature (i.e. does not do anything). Every night a batch process
shuts down Oracle instance on both the machines and copies over all the
files (Data, log, ctl etc) from Production Server to the Stand by server
(Drive to drive, directory to directory...)

In case the Production Server fails, we simply switch over the users (with 
a
different alias to the stand by server) and they are back in business.

Now, we are thinking of migrating to 8.1.7, however while trying to 
install
this version, one needs to specify a Global name which I believe has to be
unique on the network. So will the same process that I used to run (i.e.
copy all files over from production to stand by ) work??? I guess both my
servers will now have to have separate/unique Global Names. 

Is there any other approach that any of you can suggest??? 

TIA
Arif
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Re: Read Only Login with Source Visibility

2002-11-05 Thread Jared . Still
Rewrite the all_source view, and I believe the all_objects view.

It's a pain in the rear, but it can be done.

It also tends to break when you upgrade.

I did this once on Oracle 7, there may be some other workaround
now on 8i.

Jared






Bill Buchan [EMAIL PROTECTED]
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Subject:Read Only Login with Source Visibility


Hi all,

I am trying to create a read-only login, RO_USER which can do the 
following:

1. See all tables, views, constraints etc. in one  *specific* other 
schema, 
APP.

2. See all the source for PL/SQL objects in APP.

The first bit is easy: GRANT SELECT on tables, views to RO_USER.

Not sure about the second bit.  I have tried granting CREATE ANY PROCEDURE 

and CREATE ANY TRIGGER to RO_USER and using a DDL trigger to prevent these 

privileges actually being used to create procedures/triggers.  This works 
but does not restrict the source visibility to APP.  I have other schemas 
where I do not want RO_USER to see the source.

My other concern is granting SELECT on sequences as this means that they 
can select nextval from them and hence increment the numbers (not quite 
read-only!)

Any suggestions for fixing these problems would be much appreciated!

Thanks
- Bill.

PS. This is on 8i.


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BMC's dtoarchmon goes to lunch

2002-11-05 Thread Gurelei
Hi.

We are running Oracle 8172 on Dynix 452 and backing it
up via BMC's obacktrac 3.0.0.6. Since we upgraded the
OS and obakctrack dtoarchmon started dying. The
process either disappears altogether or doesn't wake
up to start an archive backup when it should. The
dtoachmon -ping shows that everything is OK. 

Has anyone experienced this and has any insight to a
possible solution?

thanks

Gene

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dispatcher timer virtual circuit status top of wait events/no M

2002-11-05 Thread Jack van Zanen
Hi All,


Above two events are at the top of my wait events in one of my databases.
Does anybody have an idea what would be the reason?

Oracle 8.1.6.0.0



Jack
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Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE

2002-11-05 Thread Jared . Still
Yechiel,

No OS errors here.  You may want to look at the
LGWR trace file in the BDUMP directory.

2 things

1) Have the SA check for device errors.  Something in the IO
system is not working.

2) Open a TAR.

If you have another location to redirect arch logs to,
you may want to do that.

The following 2 notes you will find useful for redirecting
archive logs to another location without bouncing the database.

135866.1
74324.1


HTH

Jared






Yechiel Adar [EMAIL PROTECTED]
 11/05/2002 10:56 AM

 
To: ORACLE-L [EMAIL PROTECTED], [EMAIL PROTECTED]
cc: 
Subject:Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE


Here is parts of the alert log:

  Current log# 6 seq# 4153 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1356.ORA
Sun Nov 03 11:03:34 2002
ARC0: Beginning to archive log# 5 seq# 4152
ARC0: Completed archiving log# 5 seq# 4152
Sun Nov 03 11:10:33 2002
Thread 1 advanced to log sequence 4154
  Current log# 7 seq# 4154 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1357.ORA
Sun Nov 03 11:10:33 2002
ARC2: Beginning to archive log# 6 seq# 4153
ARC2: Completed archiving log# 6 seq# 4153
Sun Nov 03 11:20:50 2002
Thread 1 advanced to log sequence 4155
  Current log# 8 seq# 4155 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1358.ORA
Sun Nov 03 11:20:50 2002
ARC1: Beginning to archive log# 7 seq# 4154
ARC1: I/O error 19502 archiving log 7 to
'E:\ORACLE\ORADATA\MUSK135\DATABASE\ARCHIVE\M135T001S04154.ARC'
ARC1: Archiving not possible: error count exceeded
ARC1: Failed to archive log# 7 seq# 4154
ARCH: Archival stopped, error occurred. Will continue retrying
ARCH:
 ORA-16038: log 7 sequence# 4154 cannot be archived
ORA-19502: write error on file , blockno  (blocksize=)
ORA-00312: online log 7 thread 1:
'D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1357.ORA'
Sun Nov 03 11:28:06 2002
Thread 1 advanced to log sequence 4156
  Current log# 3 seq# 4156 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1353.ORA
Sun Nov 03 11:28:06 2002
ARC2: Beginning to archive log# 7 seq# 4154
ARC2: Archiving not possible: No primary destinations
ARC2: Failed to archive log# 7 seq# 4154
ARCH: Archival stopped, error occurred. Will continue retrying
ARCH:
 ORA-16014: log 7 sequence# 4154 not archived, no available destinations
ORA-00312: online log 7 thread 1:
'D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1357.ORA'
Sun Nov 03 11:38:54 2002
Thread 1 advanced to log sequence 4157
  Current log# 9 seq# 4157 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOG9.ORA
Sun Nov 03 11:38:55 2002
ARC0: Beginning to archive log# 7 seq# 4154
ARC0: Archiving not possible: No primary destinations
ARC0: Failed to archive log# 7 seq# 4154
Sun Nov 03 11:39:07 2002
Thread 1 advanced to log sequence 4158
  Current log# 10 seq# 4158 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOG10.ORA
Sun Nov 03 11:39:07 2002
ARC1: Beginning to archive log# 7 seq# 4154
ARC1: Archiving not possible: No primary destinations
ARC1: Failed to archive log# 7 seq# 4154

Yechiel Adar
Mehish
- Original Message -
To: ORACLE-L [EMAIL PROTECTED]
Cc: Yechiel Adar [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 7:31 PM


 Yechiel,

 What about those alert log entries?

 Jared






 Yechiel Adar [EMAIL PROTECTED]
  11/05/2002 10:38 AM


 To: ORACLE-L [EMAIL PROTECTED],
[EMAIL PROTECTED],
 [EMAIL PROTECTED]
 cc:
 Subject:Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE
SPACE


 Hello Charlie and Jared

 Thanks for the reply.

 As I mentioned in my mail there is a lot of free space on the disk and 
the
 database is with auto archive.
 Our problem is that the ARCH process gives an error message (ora 19502)
 and
 stop. There are messages on the console about error trying to write the
 archive log, like when the disk is full. Then the redo log will fill up
 and
 the instance will stop.

 After I connect as internal and do:
 archive log stop;
 archive log all;
 archive log start;

 everything is back to normal.

 Yechiel Adar
 Mehish
 - Original Message -
 From: [EMAIL PROTECTED]
 To: [EMAIL PROTECTED]
 Sent: Tuesday, November 05, 2002 4:48 PM
 Subject: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE


  My first guess is that the instance has archiving enable, but the
 ARCHIVER
  is/was not started.
  The DB will run OK until all the redo logfiles have been filled.
  At this point the instance will simply cease to process any additional
  transactions
  until it has space to which to write additional redo logfile
 information.
 
  There should be an entry in the alert_SID.log about what is happening 

  why.
 
  - Forwarded by Charlie Mengler/THD on 11/05/2002 06:45 AM -
 
Yechiel Adar
[EMAIL PROTECTED]To:   Multiple
 recipients of list ORACLE-L [EMAIL PROTECTED]
.il cc:
Sent by: Subject:  ORA-19502 ON
 RAID
 5 DISKS WITH ENOUGH FREE SPACE
   

RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
I get an error on 8.1.7.2. Is siblings new?

SQL l
  1  SELECT LEVEL, treenode.*
  2FROM treenode
  3   START WITH parentid=0
  4  CONNECT BY PRIOR ID = parentid
  5* ORDER SIBLINGS BY PARENTid , nodeorder
SQL /
ORDER SIBLINGS BY PARENTid , nodeorder
  *
ERROR at line 5:
ORA-00924: missing BY keyword


-Original Message-
Sent: Tuesday, November 05, 2002 11:02 AM
To: '[EMAIL PROTECTED]'; Orr, Steve


SELECT LEVEL, treenode.* 
  FROM treenode 
 START WITH parentid=0 
CONNECT BY PRIOR ID = parentid 
ORDER SIBLINGS BY PARENTid , nodeorder 
Raj 
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Rajendra dot Jamadagni at ESPN dot com 
Any opinion expressed here is personal and doesn't reflect that of ESPN Inc.

QOTD: Any clod can have facts, but having an opinion is an art! 


-Original Message- 
Sent: Tuesday, November 05, 2002 12:24 PM 
To: Multiple recipients of list ORACLE-L 


Challenge: present SQL results hierarchically and sort the nodes. Use sort 
column without changing data. Here's the DDL/DML to start: 
create table treenode ( 
id  number  not null 
constraint pk_treenode primary key, 
parentidnumber  not null, 
nodeorder   number  not null, 
description varchar2(20)null); 
insert into treenode values(1,0,0,'top folder'); 
insert into treenode values(9,1,0,'1st subfolder'); 
insert into treenode values(7,1,2,'3rd subfolder'); 
insert into treenode values(2,1,1,'2nd subfolder'); 
insert into treenode values(8,7,1,'folder 3 item 2'); 
insert into treenode values(6,2,3,'folder 2 item 3'); 
insert into treenode values(5,7,0,'folder 3 item 1'); 
insert into treenode values(3,2,2,'folder 2 item 2'); 
insert into treenode values(4,2,1,'folder 2 item 1'); 
- 
Here's the data presented hierachically without the desired sort: 
select * from treenode 
start with parentid=0 connect by prior id = parentid; 
ID   PARENTID  NODEORDER DESCRIPTION 
-- -- --  
 1  0  0 top folder 
 9  1  0 1st subfolder 
 7  1  2 3rd subfolder 
 8  7  1 folder 3 item 2 
 5  7  0 folder 3 item 1 
 2  1  1 2nd subfolder 
 6  2  3 folder 2 item 3 
 3  2  2 folder 2 item 2 
 4  2  1 folder 2 item 1 
- 
Desired SQL statement results: 
ID   PARENTID  NODEORDER DESCRIPTION 
-- -- --  
 1  0  0 top folder 
 9  1  0 1st subfolder 
 2  1  1 2nd subfolder 
 4  2  1 folder 2 item 1 
 3  2  2 folder 2 item 2 
 6  2  3 folder 2 item 3 
 7  1  2 3rd subfolder 
 5  7  0 folder 3 item 1 
 8  7  1 folder 3 item 2 
- 
Kudos to anyone who can figure out how to do this via SQL. 


Steve Orr 
Bozeman, Montana 
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Re: ORA-02050

2002-11-05 Thread Yechiel Adar
There is another view, DBA_2PC_NEIGHBORS.
Maybe there is some info there.

Yechiel Adar
Mehish
- Original Message -
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 5:58 PM


 On Tue, Nov 05, 2002 at 06:33:44AM -0800, [EMAIL PROTECTED]
wrote:
  Ray,
  I think you need to check DBA_2PC_PENDING on both the local and remote
  database to see if anything is still in there.
  That gives the osuser, terminal and host name so you may be able to get
  something from that.
  Of course if there is nothing in that table then the transaction has
been
  rolled back already


 Right, nothing there, so I was wondering if there was a way to chase
 the transaction id to app, such as archivelog.   8.1.7.4 here.

 Thanks.





  John
 
  -Original Message-
  Sent: 05 November 2002 13:43
  To: Multiple recipients of list ORACLE-L
 
 
 
  ORA-02050: transaction 8.82.26033 rolled back, some remote DBs may be
  in-doubt
 
  The transaction seems to have completely rolled back.  My
  question is, is there any way to relate transaction id 8.82.26033
  to an application or table row or something at a higher layer?
 
  Thanks.
  ===
  Ray Stell   [EMAIL PROTECTED] (540) 231-4109 KE4TJC28^D
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  Author: Ray Stell
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Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE

2002-11-05 Thread Jared . Still
Yechiel,

What about those alert log entries?

Jared






Yechiel Adar [EMAIL PROTECTED]
 11/05/2002 10:38 AM

 
To: ORACLE-L [EMAIL PROTECTED], [EMAIL PROTECTED], 
[EMAIL PROTECTED]
cc: 
Subject:Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE


Hello Charlie and Jared

Thanks for the reply.

As I mentioned in my mail there is a lot of free space on the disk and the
database is with auto archive.
Our problem is that the ARCH process gives an error message (ora 19502) 
and
stop. There are messages on the console about error trying to write the
archive log, like when the disk is full. Then the redo log will fill up 
and
the instance will stop.

After I connect as internal and do:
archive log stop;
archive log all;
archive log start;

everything is back to normal.

Yechiel Adar
Mehish
- Original Message -
To: [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 4:48 PM


 My first guess is that the instance has archiving enable, but the 
ARCHIVER
 is/was not started.
 The DB will run OK until all the redo logfiles have been filled.
 At this point the instance will simply cease to process any additional
 transactions
 until it has space to which to write additional redo logfile 
information.

 There should be an entry in the alert_SID.log about what is happening 
 why.

 - Forwarded by Charlie Mengler/THD on 11/05/2002 06:45 AM -

   Yechiel Adar
   [EMAIL PROTECTED]To:   Multiple
recipients of list ORACLE-L [EMAIL PROTECTED]
   .il cc:
   Sent by: Subject:  ORA-19502 ON 
RAID
5 DISKS WITH ENOUGH FREE SPACE
   [EMAIL PROTECTED]



   11/05/2002 12:58
   AM
   Please respond to
   ORACLE-L






 Hello All

 Oracle 8.1.6.3.4 on NT.

 I got ora-19502 - can not write archive log, and users could not logged
on.
 I can logon only as internal.
 We had the same problem a few days ago.
 The technical support people checked the raid 5 disks and did not find 
any
 I/O errors.
 There is a lot of free space on the drive.

 Has anybody encountered this problem?
 Do you know what can cause it?
 What I need to do in order to prevent this from happening again?


 Yechiel Adar
 Mehish

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 Author: Yechiel Adar
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Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE

2002-11-05 Thread Yechiel Adar
Thanks Jared.

I forgot to mention that the SA checked for device errors. None found.

A tar has been opened is already on it's way to higher support level.

This problem occurred again today, third time in two weeks.

Also, the database is up, only the ARCH process is dead.
I started the ARCH (archive log stop; archive log all; archive log start;)
and the database continued to work.

Yechiel Adar
Mehish
- Original Message -
To: ORACLE-L [EMAIL PROTECTED]
Cc: Yechiel Adar [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 8:17 PM


 Yechiel,

 No OS errors here.  You may want to look at the
 LGWR trace file in the BDUMP directory.

 2 things

 1) Have the SA check for device errors.  Something in the IO
 system is not working.

 2) Open a TAR.

 If you have another location to redirect arch logs to,
 you may want to do that.

 The following 2 notes you will find useful for redirecting
 archive logs to another location without bouncing the database.

 135866.1
 74324.1


 HTH

 Jared








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Re: SQL Brain Teaser Challenge

2002-11-05 Thread Jay Hostetter
Does an in-line view do the trick?
select * from
(select id, parentid, nodeorder, description
from treenode
start with parentid=0 connect by prior id = parentid)
order by nodeorder;

Jay

 [EMAIL PROTECTED] 11/05/02 12:24PM 
Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
id  number  not null 
constraint pk_treenode primary key,
parentidnumber  not null,
nodeorder   number  not null,
description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2
-

Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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**DISCLAIMER
This e-mail message and any files transmitted with it are intended for the use of the 
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may not use, copy or disclose to anyone the message or any information contained in 
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RE: Read Only Login with Source Visibility

2002-11-05 Thread Mercadante, Thomas F
better yet - create a local copy of it, and place a private synonym in their
account to point to it.

really sneaky.


-Original Message-
Sent: Tuesday, November 05, 2002 12:24 PM
To: Multiple recipients of list ORACLE-L


Rewrite the all_source view, and I believe the all_objects view.

It's a pain in the rear, but it can be done.

It also tends to break when you upgrade.

I did this once on Oracle 7, there may be some other workaround
now on 8i.

Jared






Bill Buchan [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 04:48 AM
 Please respond to ORACLE-L

 
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
cc: 
Subject:Read Only Login with Source Visibility


Hi all,

I am trying to create a read-only login, RO_USER which can do the 
following:

1. See all tables, views, constraints etc. in one  *specific* other 
schema, 
APP.

2. See all the source for PL/SQL objects in APP.

The first bit is easy: GRANT SELECT on tables, views to RO_USER.

Not sure about the second bit.  I have tried granting CREATE ANY PROCEDURE 

and CREATE ANY TRIGGER to RO_USER and using a DDL trigger to prevent these 

privileges actually being used to create procedures/triggers.  This works 
but does not restrict the source visibility to APP.  I have other schemas 
where I do not want RO_USER to see the source.

My other concern is granting SELECT on sequences as this means that they 
can select nextval from them and hence increment the numbers (not quite 
read-only!)

Any suggestions for fixing these problems would be much appreciated!

Thanks
- Bill.

PS. This is on 8i.


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RE: OCP Exams - What to study out of?

2002-11-05 Thread Paula_Stankus
Title: RE: OCP Exams - What to study out of?





I took the SQL exam first and had no problems although I have heard others have. I think it has to do with how I learned - I forced myself to use SQL before getting hooked into a GUI tool of any kind. In the long-run that is a good approach as you will always have that but might not always have the GUI.

-Original Message-
From: DENNIS WILLIAMS [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, November 05, 2002 12:04 PM
To: Multiple recipients of list ORACLE-L
Subject: RE: OCP Exams - What to study out of?



Eva - Everything I've heard says that all the exam questions come out of the
Oracle University class Student Guides. To put it bluntly, your goal is to
pass the exam. You could study the Oracle manuals, but there is a lot of
material to cover and what strikes you as important might not be covered on
the exam, and vice-versa. Essentially you are getting two levels of
filtering. Somebody read the manuals and used that to prepare student
guides, then a group of Oracle instructors read the student guide and
prepared questions. I think that most of the exam guides were prepared from
the Oracle Student Guides and the author has at least taken the exam, so
they have a general idea of how the test is approached. Everyone has their
favorite exam preparation book and I consider this a worthwhile investment,
I bought Couchman -
http://search.barnesandnoble.com/booksearch/isbnInquiry.asp?userid=6WIANMIL0
H
http://search.barnesandnoble.com/booksearch/isbnInquiry.asp?userid=6WIANMIL
0Hsourceid=00400731866779927243bfdate=11%2D05%2D2002+10%3A16%3A17isbn=007
2133414
sourceid=00400731866779927243bfdate=11%2D05%2D2002+10%3A16%3A17isbn=00721
33414
 Two issues: Which exam track are you planning to take? 9i? 8i? 
 The advice I received is don't take the SQL exam as your first exam.
Take the DBA exam first. The SQL exam is reported to be quite hard because
it covers all the odd SQL functions and can really catch you unawares. As a
practicing DBA you should find the DBA exam pretty easy. I didn't personally
take that path for my own reasons, but I still consider it good advice.




Dennis Williams 
DBA, 40%OCP 
Lifetouch, Inc. 
[EMAIL PROTECTED] 


-Original Message-
Sent: Tuesday, November 05, 2002 6:09 AM
To: Multiple recipients of list ORACLE-L




Hello Everyone, 


1. I have now made the humungous decision to start studying and to write the
OCP exams. 
2. Do I study out of the Oracle Manuals? 
3. I do have the Sybex Study Guides, would studying these be all that is
needed? 
4. Or do the questions come out of the Course material ( Which I have not
attended ). 
5. I do have some of the Oreilly Insect Books - would these be any help? 


Any views, opinions etc appreciated. 
Regards 
Denham Eva 
Oracle DBA 
UNIX is basically a simple operating system, but you have to be a genius to
understand the simplicity. 
Dennis Ritchie. 




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Re: SQL Brain Teaser Challenge

2002-11-05 Thread Jared . Still
Steve,

This works for me.

Jared



col nodelevel noprint
col parent noprint
col child noprint

select
   a.nodelevel
   , a.id id
   , a.parentid
   , a.nodeorder
   , a.description
   , decode(c.children,null,'N','Y') parent
   , decode(p.children,null,'Y','N') child
from (
   select level nodelevel, id, parentid, nodeorder, description
   from treenode
   start with parentid=0
   connect by prior id = parentid
) a,
(
   select parentid, count(*) children
   from treenode b
   group by parentid
) c,
(
   select parentid, count(*) children
   from treenode d
   group by parentid
) p
where a.id = c.parentid(+)
and a.id = p.parentid(+)
order by
   decode(parent
  -- is a parent
  , 'Y', nodelevel * id
  -- is a child
  , 'N', nodelevel * parentid + nodeorder
   )
/






Orr, Steve [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 09:24 AM
 Please respond to ORACLE-L

 
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
cc: 
Subject:SQL Brain Teaser Challenge


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
 id  number  not null 
 constraint pk_treenode 
primary key,
 parentidnumber  not null,
 nodeorder   number  not null,
 description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2
-

Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Whittle Jerome Contr NCI
Title: RE: SQL Brain Teaser Challenge






Steve,


I'm busted! You caught me cheating. Still you must admit it was an interesting use of TRANSLATE given the data in the DESCRIPTION field.

Jerry Whittle

ACIFICS DBA

NCI Information Systems Inc.

[EMAIL PROTECTED]

618-622-4145


-Original Message-

From: Orr, Steve [SMTP:[EMAIL PROTECTED]]


Well it worked but it doesn't follow the rules. The description can change

so it should not be sorted on. Consider this:


update treenode set description='2nd item, 3rd folder' where id=8;


select * from treenode

order by translate(description || 0, 

 '0123456789abcdefghijklmnopqrstuvwxyz ', '0123456789');


 ID PARENTID NODEORDER DESCRIPTION

-- -- -- 

 1 0 0 top folder

 9 1 0 1st subfolder

 8 7 1 2nd item, 3rd folder

 2 1 1 2nd subfolder

 4 2 1 folder 2 item 1

 3 2 2 folder 2 item 2

 6 2 3 folder 2 item 3

 7 1 2 3rd subfolder

 5 7 0 folder 3 item 1


9 rows selected.

-Original Message-

From: Whittle Jerome Contr NCI [mailto:[EMAIL PROTECTED]]


select * 

from treenode 

order by translate(description || 0, '0123456789abcdefghijklmnopqrstuvwxyz

', '0123456789') 

Jerry Whittle 

ACIFICS DBA 

NCI Information Systems Inc. 

[EMAIL PROTECTED] 

618-622-4145 

-Original Message- 

From: Orr, Steve [SMTP:[EMAIL PROTECTED]] 

Challenge: present SQL results hierarchically and sort the nodes. Use sort 

column without changing data. Here's the DDL/DML to start: 

create table treenode ( 

 id number not null 

 constraint pk_treenode primary key, 

 parentid number not null, 

 nodeorder number not null, 

 description varchar2(20) null); 

insert into treenode values(1,0,0,'top folder'); 

insert into treenode values(9,1,0,'1st subfolder'); 

insert into treenode values(7,1,2,'3rd subfolder'); 

insert into treenode values(2,1,1,'2nd subfolder'); 

insert into treenode values(8,7,1,'folder 3 item 2'); 

insert into treenode values(6,2,3,'folder 2 item 3'); 

insert into treenode values(5,7,0,'folder 3 item 1'); 

insert into treenode values(3,2,2,'folder 2 item 2'); 

insert into treenode values(4,2,1,'folder 2 item 1'); 

- 

Here's the data presented hierachically without the desired sort: 

select * from treenode 

start with parentid=0 connect by prior id = parentid; 

 ID PARENTID NODEORDER DESCRIPTION 

-- -- --  

 1 0 0 top folder 

 9 1 0 1st subfolder 

 7 1 2 3rd subfolder 

 8 7 1 folder 3 item 2 

 5 7 0 folder 3 item 1 

 2 1 1 2nd subfolder 

 6 2 3 folder 2 item 3 

 3 2 2 folder 2 item 2 

 4 2 1 folder 2 item 1 

- 

Desired SQL statement results: 

 ID PARENTID NODEORDER DESCRIPTION 

-- -- --  

 1 0 0 top folder 

 9 1 0 1st subfolder 

 2 1 1 2nd subfolder 

 4 2 1 folder 2 item 1 

 3 2 2 folder 2 item 2 

 6 2 3 folder 2 item 3 

 7 1 2 3rd subfolder 

 5 7 0 folder 3 item 1 

 8 7 1 folder 3 item 2 

- 

Kudos to anyone who can figure out how to do this via SQL. 

Steve Orr 

Bozeman, Montana 





Re: SQL Brain Teaser Challenge

2002-11-05 Thread Jared . Still
Steve,

This works for me.

Jared



col nodelevel noprint
col parent noprint
col child noprint

select
   a.nodelevel
   , a.id id
   , a.parentid
   , a.nodeorder
   , a.description
   , decode(c.children,null,'N','Y') parent
   , decode(p.children,null,'Y','N') child
from (
   select level nodelevel, id, parentid, nodeorder, description
   from treenode
   start with parentid=0
   connect by prior id = parentid
) a,
(
   select parentid, count(*) children
   from treenode b
   group by parentid
) c,
(
   select parentid, count(*) children
   from treenode d
   group by parentid
) p
where a.id = c.parentid(+)
and a.id = p.parentid(+)
order by
   decode(parent
  -- is a parent
  , 'Y', nodelevel * id
  -- is a child
  , 'N', nodelevel * parentid + nodeorder
   )
/






Orr, Steve [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 09:24 AM
 Please respond to ORACLE-L

 
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
cc: 
Subject:SQL Brain Teaser Challenge


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
 id  number  not null 
 constraint pk_treenode 
primary key,
 parentidnumber  not null,
 nodeorder   number  not null,
 description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2
-

Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE

2002-11-05 Thread Jared . Still
Did I say LGWR trace file?

I meant ARCH trace file, sorry.

jared






[EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 10:23 AM
 Please respond to ORACLE-L

 
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
cc: 
Subject:Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE


Yechiel,

No OS errors here.  You may want to look at the
LGWR trace file in the BDUMP directory.

2 things

1) Have the SA check for device errors.  Something in the IO
system is not working.

2) Open a TAR.

If you have another location to redirect arch logs to,
you may want to do that.

The following 2 notes you will find useful for redirecting
archive logs to another location without bouncing the database.

135866.1
74324.1


HTH

Jared






Yechiel Adar [EMAIL PROTECTED]
 11/05/2002 10:56 AM

 
To: ORACLE-L [EMAIL PROTECTED], 
[EMAIL PROTECTED]
cc: 
Subject:Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE 
SPACE


Here is parts of the alert log:

  Current log# 6 seq# 4153 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1356.ORA
Sun Nov 03 11:03:34 2002
ARC0: Beginning to archive log# 5 seq# 4152
ARC0: Completed archiving log# 5 seq# 4152
Sun Nov 03 11:10:33 2002
Thread 1 advanced to log sequence 4154
  Current log# 7 seq# 4154 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1357.ORA
Sun Nov 03 11:10:33 2002
ARC2: Beginning to archive log# 6 seq# 4153
ARC2: Completed archiving log# 6 seq# 4153
Sun Nov 03 11:20:50 2002
Thread 1 advanced to log sequence 4155
  Current log# 8 seq# 4155 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1358.ORA
Sun Nov 03 11:20:50 2002
ARC1: Beginning to archive log# 7 seq# 4154
ARC1: I/O error 19502 archiving log 7 to
'E:\ORACLE\ORADATA\MUSK135\DATABASE\ARCHIVE\M135T001S04154.ARC'
ARC1: Archiving not possible: error count exceeded
ARC1: Failed to archive log# 7 seq# 4154
ARCH: Archival stopped, error occurred. Will continue retrying
ARCH:
 ORA-16038: log 7 sequence# 4154 cannot be archived
ORA-19502: write error on file , blockno  (blocksize=)
ORA-00312: online log 7 thread 1:
'D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1357.ORA'
Sun Nov 03 11:28:06 2002
Thread 1 advanced to log sequence 4156
  Current log# 3 seq# 4156 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1353.ORA
Sun Nov 03 11:28:06 2002
ARC2: Beginning to archive log# 7 seq# 4154
ARC2: Archiving not possible: No primary destinations
ARC2: Failed to archive log# 7 seq# 4154
ARCH: Archival stopped, error occurred. Will continue retrying
ARCH:
 ORA-16014: log 7 sequence# 4154 not archived, no available destinations
ORA-00312: online log 7 thread 1:
'D:\ORACLE\ORADATA\MUSK135\DATABASE\LOGM1357.ORA'
Sun Nov 03 11:38:54 2002
Thread 1 advanced to log sequence 4157
  Current log# 9 seq# 4157 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOG9.ORA
Sun Nov 03 11:38:55 2002
ARC0: Beginning to archive log# 7 seq# 4154
ARC0: Archiving not possible: No primary destinations
ARC0: Failed to archive log# 7 seq# 4154
Sun Nov 03 11:39:07 2002
Thread 1 advanced to log sequence 4158
  Current log# 10 seq# 4158 mem# 0:
D:\ORACLE\ORADATA\MUSK135\DATABASE\LOG10.ORA
Sun Nov 03 11:39:07 2002
ARC1: Beginning to archive log# 7 seq# 4154
ARC1: Archiving not possible: No primary destinations
ARC1: Failed to archive log# 7 seq# 4154

Yechiel Adar
Mehish
- Original Message -
To: ORACLE-L [EMAIL PROTECTED]
Cc: Yechiel Adar [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 7:31 PM


 Yechiel,

 What about those alert log entries?

 Jared






 Yechiel Adar [EMAIL PROTECTED]
  11/05/2002 10:38 AM


 To: ORACLE-L [EMAIL PROTECTED],
[EMAIL PROTECTED],
 [EMAIL PROTECTED]
 cc:
 Subject:Re: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE
SPACE


 Hello Charlie and Jared

 Thanks for the reply.

 As I mentioned in my mail there is a lot of free space on the disk and 
the
 database is with auto archive.
 Our problem is that the ARCH process gives an error message (ora 19502)
 and
 stop. There are messages on the console about error trying to write the
 archive log, like when the disk is full. Then the redo log will fill up
 and
 the instance will stop.

 After I connect as internal and do:
 archive log stop;
 archive log all;
 archive log start;

 everything is back to normal.

 Yechiel Adar
 Mehish
 - Original Message -
 From: [EMAIL PROTECTED]
 To: [EMAIL PROTECTED]
 Sent: Tuesday, November 05, 2002 4:48 PM
 Subject: ORA-19502 ON RAID 5 DISKS WITH ENOUGH FREE SPACE


  My first guess is that the instance has archiving enable, but the
 ARCHIVER
  is/was not started.
  The DB will run OK until all the redo logfiles have been filled.
  At this point the instance will simply cease to process any additional
  transactions
  until it has space to which to write additional redo logfile
 information.
 
  There should be an entry in the alert_SID.log about what is happening 

  why.
 
  - Forwarded by Charlie Mengler/THD on 11/05/2002 

RE: SQL Brain Teaser Challenge

2002-11-05 Thread Fink, Dan
I love a good challenge. Since you cannot sort on a hierarchical query, you
have to use an inline query...

select id, parentid, nodeorder, description
from (select id, parentid, nodeorder, description from treenod
  start with parentid=0 connect by prior id = parentid)
order by parentid, nodeorder;

   ID   PARENTID  NODEORDER DESCRIPTION
- -- -- 
1  0  0 top folder
9  1  0 1st subfolder
2  1  1 2nd subfolder
7  1  2 3rd subfolder
4  2  1 folder 2 item 1
3  2  2 folder 2 item 2
6  2  3 folder 2 item 3
5  7  0 folder 3 item 1
8  7  1 folder 3 item 2

Dan Fink

-Original Message-
Sent: Tuesday, November 05, 2002 10:24 AM
To: Multiple recipients of list ORACLE-L


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
id  number  not null 
constraint pk_treenode primary key,
parentidnumber  not null,
nodeorder   number  not null,
description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2
-

Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Whittle Jerome Contr NCI
Title: RE: SQL Brain Teaser Challenge






select *

from treenode 

order by translate(description || 0, '0123456789abcdefghijklmnopqrstuvwxyz ', '0123456789') 


Jerry Whittle

ACIFICS DBA

NCI Information Systems Inc.

[EMAIL PROTECTED]

618-622-4145


-Original Message-

From: Orr, Steve [SMTP:[EMAIL PROTECTED]]


Challenge: present SQL results hierarchically and sort the nodes. Use sort

column without changing data. Here's the DDL/DML to start:


create table treenode (

 id  number  not null 

   constraint pk_treenode primary key,

 parentid number  not null,

 nodeorder number  not null,

 description varchar2(20)  null);


insert into treenode values(1,0,0,'top folder');

insert into treenode values(9,1,0,'1st subfolder');

insert into treenode values(7,1,2,'3rd subfolder');

insert into treenode values(2,1,1,'2nd subfolder');

insert into treenode values(8,7,1,'folder 3 item 2');

insert into treenode values(6,2,3,'folder 2 item 3');

insert into treenode values(5,7,0,'folder 3 item 1');

insert into treenode values(3,2,2,'folder 2 item 2');

insert into treenode values(4,2,1,'folder 2 item 1');

-

Here's the data presented hierachically without the desired sort:

select * from treenode 

start with parentid=0 connect by prior id = parentid;

 ID PARENTID NODEORDER DESCRIPTION

-- -- -- 

 1 0 0 top folder

 9 1 0 1st subfolder

 7 1 2 3rd subfolder

 8 7 1 folder 3 item 2

 5 7 0 folder 3 item 1

 2 1 1 2nd subfolder

 6 2 3 folder 2 item 3

 3 2 2 folder 2 item 2

 4 2 1 folder 2 item 1

-

Desired SQL statement results:

 ID PARENTID NODEORDER DESCRIPTION

-- -- -- 

 1 0 0 top folder

 9 1 0 1st subfolder

 2 1 1 2nd subfolder

 4 2 1 folder 2 item 1

 3 2 2 folder 2 item 2

 6 2 3 folder 2 item 3

 7 1 2 3rd subfolder

 5 7 0 folder 3 item 1

 8 7 1 folder 3 item 2

-


Kudos to anyone who can figure out how to do this via SQL.


Steve Orr

Bozeman, Montana





RE: SQL Brain Teaser Challenge

2002-11-05 Thread Mirsky, Greg
How about...

SELECT t.id
 , t.parentid
 , t.nodeorder
 , t.description
  FROM treenode t
CONNECT BY t.parentid = PRIOR t.id
START WITH t.description = 'top folder'
  ORDER BY NVL (
  TRIM (
 TRANSLATE (
LOWER (t.description)
  , 'abcdefghijklmnopqrstuvwxyz'
  , '  '
 )
  )
, 0
   );


Regards!

Greg

-Original Message-
Sent: Tuesday, November 05, 2002 12:24 PM
To: Multiple recipients of list ORACLE-L


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
id  number  not null 
constraint pk_treenode primary key,
parentidnumber  not null,
nodeorder   number  not null,
description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2
-

Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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Please see the official ORACLE-L FAQ: http://www.orafaq.com
-- 
Author: Orr, Steve
  INET: [EMAIL PROTECTED]

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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Jamadagni, Rajendra
Title: RE: SQL Brain Teaser Challenge





Steve,
I did try it ...


oraclei@rhea-ACPT1 sys


SQL*Plus: Release 9.2.0.1.0 - Production on Tue Nov 5 14:18:02 2002
Copyright (c) 1982, 2002, Oracle Corporation. All rights reserved.
Connected.
SQL conn system
Enter password: 
Connected.
SQL SELECT *
 FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder) 
START WITH parentid=0 
CONNECT BY PRIOR ID = parentid
SQL /


 ID PARENTID NODEORDER DESCRIPTION
-- -- -- 
 1 0 0 top folder
 9 1 0 1st subfolder
 2 1 1 2nd subfolder
 4 2 1 folder 2 item 1
 3 2 2 folder 2 item 2
 6 2 3 folder 2 item 3
 7 1 2 3rd subfolder
 5 7 0 folder 3 item 1
 8 7 1 folder 3 item 2


9 rows selected.



Raj
__
Rajendra Jamadagni  MIS, ESPN Inc.
Rajendra dot Jamadagni at ESPN dot com
Any opinion expressed here is personal and doesn't reflect that of ESPN Inc. 
QOTD: Any clod can have facts, but having an opinion is an art!



-Original Message-
From: Orr, Steve [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, November 05, 2002 2:15 PM
To: Jamadagni, Rajendra; '[EMAIL PROTECTED]'
Subject: RE: SQL Brain Teaser Challenge



Nope...


SQL l
 1 SELECT *
 2 FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder)
 3 START WITH parentid=0
 4* CONNECT BY PRIOR ID = parentid
SQL /
 FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder)
 *
ERROR at line 2:
ORA-01472: cannot use CONNECT BY on view with DISTINCT, GROUP BY, etc.



I supplied the DDL/DML so you could test on your own systems.




-Original Message-
From: Jamadagni, Rajendra [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, November 05, 2002 12:07 PM
To: '[EMAIL PROTECTED]'; Orr, Steve
Subject: RE: SQL Brain Teaser Challenge



SELECT * 
 FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder) 
START WITH parentid=0 
CONNECT BY PRIOR ID = parentid 
Does this work? 
Also see
http://asktom.oracle.com/pls/ask/f?p=4950:8:F4950_P8_DISPLAYID:921234804
9 
Raj 
__ 
Rajendra Jamadagni MIS, ESPN Inc. 
Rajendra dot Jamadagni at ESPN dot com 
Any opinion expressed here is personal and doesn't reflect that of ESPN Inc.


QOTD: Any clod can have facts, but having an opinion is an art! 



This e-mail 
message is confidential, intended only for the named recipient(s) above and may 
contain information that is privileged, attorney work product or exempt from 
disclosure under applicable law. If you have received this message in error, or are 
not the named recipient(s), please immediately notify corporate MIS at (860) 766-2000 
and delete this e-mail message from your computer, Thank 
you.*2



RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
Title: RE: SQL Brain Teaser Challenge



Whoa... A query that works on 9i but fails on 8i 
!!!


  -Original Message-From: Jamadagni, Rajendra 
  [mailto:[EMAIL PROTECTED]]Sent: Tuesday, November 05, 
  2002 12:19 PMTo: Orr, Steve; 
  '[EMAIL PROTECTED]'Subject: RE: SQL Brain Teaser 
  Challenge
  Steve, I did try it ... 
  oraclei@rhea-ACPT1 sys 
  SQL*Plus: Release 9.2.0.1.0 - Production on Tue Nov 5 14:18:02 
  2002 Copyright (c) 1982, 2002, Oracle 
  Corporation. All rights reserved. Connected. SQL conn system Enter password: Connected. SQL SELECT *  FROM (SELECT * FROM 
  treenode ORDER BY parentid, nodeorder) START WITH 
  parentid=0 CONNECT BY PRIOR ID = parentid 
  SQL / 
   ID 
  PARENTID NODEORDER DESCRIPTION -- 
  -- --   
  1 
  0 0 top folder 
   
  9 
  1 0 1st subfolder 
   
  2 
  1 1 2nd subfolder 
   
  4 
  2 1 folder 2 item 
  1  
  3 
  2 2 folder 2 item 
  2  
  6 
  2 3 folder 2 item 
  3  
  7 
  1 2 3rd subfolder 
   
  5 
  7 0 folder 3 item 
  1  
  8 
  7 1 folder 3 item 
  2 
  9 rows selected. 
  Raj __ Rajendra Jamadagni 
   MIS, ESPN Inc. Rajendra dot Jamadagni at ESPN dot com Any 
  opinion expressed here is personal and doesn't reflect that of ESPN Inc. 
  QOTD: Any clod can have facts, but having an opinion 
  is an art! 
  -Original Message- From: Orr, 
  Steve [mailto:[EMAIL PROTECTED]] 
  Sent: Tuesday, November 05, 2002 2:15 PM To: Jamadagni, Rajendra; '[EMAIL PROTECTED]' Subject: RE: SQL Brain Teaser Challenge 
  Nope... 
  SQL l  1 SELECT 
  *  2 FROM (SELECT * FROM 
  treenode ORDER BY parentid, nodeorder)  3 
  START WITH parentid=0  4* CONNECT BY PRIOR ID = 
  parentid SQL /  FROM 
  (SELECT * FROM treenode ORDER BY parentid, nodeorder)  
  * ERROR at line 2: ORA-01472: 
  cannot use CONNECT BY on view with DISTINCT, GROUP BY, etc. 
  I supplied the DDL/DML so you could test on your own 
  systems. 
  -Original Message- From: 
  Jamadagni, Rajendra [mailto:[EMAIL PROTECTED]] 
  Sent: Tuesday, November 05, 2002 12:07 PM To: '[EMAIL PROTECTED]'; Orr, Steve Subject: 
  RE: SQL Brain Teaser Challenge 
  SELECT *  FROM (SELECT * FROM 
  treenode ORDER BY parentid, nodeorder) START WITH 
  parentid=0 CONNECT BY PRIOR ID = parentid 
  Does this work? Also 
  see http://asktom.oracle.com/pls/ask/f?p=4950:8:F4950_P8_DISPLAYID:921234804 
  9 Raj __ Rajendra 
  Jamadagni 
  MIS, ESPN Inc. Rajendra dot Jamadagni at ESPN dot com 
  Any opinion expressed here is personal and doesn't 
  reflect that of ESPN Inc. 
  QOTD: Any clod can have facts, but having an opinion is an 
  art! 


RE: OCP Exams - What to study out of?

2002-11-05 Thread Rick_Cale

The Cloverleaf product I know is from Quovadx.  One of its usages it
provides an interface between Healthcare systems and Oracle(many  other
DBMS).  For ex. we use to interface with hospitals sending data in standard
HL7 format and Cloverleaf interprets
HL7 format and data is loaded directly into Oracle for further processing.

See http://www.quovadx.com/ for more info

Rick




   

[EMAIL PROTECTED]

tate.fl.usTo: Multiple recipients of list 
ORACLE-L 
Sent by:   [EMAIL PROTECTED]  

[EMAIL PROTECTED]  cc:  

  Subject: RE: OCP Exams - What to 
study out of?   
   

11/05/2002 10:23 AM

Please respond to  

ORACLE-L   

   

   





Guys, anyone know about an application called Cloverleaf?  Basically, my
organization is pursuing using Cloverleaf as the data integration sole
application of choice.  To me Unix/SQL*Loader and native tools for the
various RDBMS usually are more straightforward and provide simplication and
less proprietary coding and it is easier for an organization to find others
who have those type of skills.  Any ideas guys?


-Original Message-
Sent: Tuesday, November 05, 2002 9:18 AM
To: Multiple recipients of list ORACLE-L





Denham,
I have just passed my 8i upgrade exam and added some notes to a page about
getting OCP certification.
It can be accessed on http://www.hcresources.co.uk/ocp.htm
http://www.hcresources.co.uk/ocp.htm

Hope you find it interesting

John


-Original Message-
Sent: 05 November 2002 12:09
To: Multiple recipients of list ORACLE-L






Hello Everyone,


1. I have now made the humungous decision to start studying and to write
the
OCP exams.
2. Do I study out of the Oracle Manuals?
3. I do have the Sybex Study Guides, would studying these be all that is
needed?
4. Or do the questions come out of the Course material ( Which I have not
attended ).
5. I do have some of the Oreilly Insect Books - would these be any help?


Any views, opinions etc appreciated.
Regards
Denham Eva
Oracle DBA
UNIX is basically a simple operating system, but you have to be a genius
to
understand the simplicity.
Dennis Ritchie.






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Fat City 

RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
You still haven't answered the challenge... your results are in a different
order than the desired results I gave at the bottom. The idea is to have the
children appearing immediately after the parents in the correct order.



-Original Message-
Sent: Tuesday, November 05, 2002 11:57 AM
To: '[EMAIL PROTECTED]'; Orr, Steve


I love a good challenge. Since you cannot sort on a hierarchical query, you
have to use an inline query...

select id, parentid, nodeorder, description
from (select id, parentid, nodeorder, description from treenod
  start with parentid=0 connect by prior id = parentid)
order by parentid, nodeorder;

   ID   PARENTID  NODEORDER DESCRIPTION
- -- -- 
1  0  0 top folder
9  1  0 1st subfolder
2  1  1 2nd subfolder
7  1  2 3rd subfolder
4  2  1 folder 2 item 1
3  2  2 folder 2 item 2
6  2  3 folder 2 item 3
5  7  0 folder 3 item 1
8  7  1 folder 3 item 2

Dan Fink

-Original Message-
Sent: Tuesday, November 05, 2002 10:24 AM
To: Multiple recipients of list ORACLE-L


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
id  number  not null 
constraint pk_treenode primary key,
parentidnumber  not null,
nodeorder   number  not null,
description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2
-

Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
Well it worked but it doesn't follow the rules. The description can change
so it should not be sorted on. Consider this:

update treenode set description='2nd item, 3rd folder' where id=8;

select * from treenode
order by translate(description || 0, 
'0123456789abcdefghijklmnopqrstuvwxyz ', '0123456789');

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 8  7  1 2nd item, 3rd folder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1

9 rows selected.
-Original Message-
Sent: Tuesday, November 05, 2002 11:52 AM
To: [EMAIL PROTECTED]
Cc: Orr, Steve


select * 
from treenode 
order by translate(description || 0, '0123456789abcdefghijklmnopqrstuvwxyz
', '0123456789') 
Jerry Whittle 
ACIFICS DBA 
NCI Information Systems Inc. 
[EMAIL PROTECTED] 
618-622-4145 
-Original Message- 
Challenge: present SQL results hierarchically and sort the nodes. Use sort 
column without changing data. Here's the DDL/DML to start: 
create table treenode ( 
id  number  not null 
constraint pk_treenode primary key, 
parentidnumber  not null, 
nodeorder   number  not null, 
description varchar2(20)null); 
insert into treenode values(1,0,0,'top folder'); 
insert into treenode values(9,1,0,'1st subfolder'); 
insert into treenode values(7,1,2,'3rd subfolder'); 
insert into treenode values(2,1,1,'2nd subfolder'); 
insert into treenode values(8,7,1,'folder 3 item 2'); 
insert into treenode values(6,2,3,'folder 2 item 3'); 
insert into treenode values(5,7,0,'folder 3 item 1'); 
insert into treenode values(3,2,2,'folder 2 item 2'); 
insert into treenode values(4,2,1,'folder 2 item 1'); 
- 
Here's the data presented hierachically without the desired sort: 
select * from treenode 
start with parentid=0 connect by prior id = parentid; 
ID   PARENTID  NODEORDER DESCRIPTION 
-- -- --  
 1  0  0 top folder 
 9  1  0 1st subfolder 
 7  1  2 3rd subfolder 
 8  7  1 folder 3 item 2 
 5  7  0 folder 3 item 1 
 2  1  1 2nd subfolder 
 6  2  3 folder 2 item 3 
 3  2  2 folder 2 item 2 
 4  2  1 folder 2 item 1 
- 
Desired SQL statement results: 
ID   PARENTID  NODEORDER DESCRIPTION 
-- -- --  
 1  0  0 top folder 
 9  1  0 1st subfolder 
 2  1  1 2nd subfolder 
 4  2  1 folder 2 item 1 
 3  2  2 folder 2 item 2 
 6  2  3 folder 2 item 3 
 7  1  2 3rd subfolder 
 5  7  0 folder 3 item 1 
 8  7  1 folder 3 item 2 
- 
Kudos to anyone who can figure out how to do this via SQL. 
Steve Orr 
Bozeman, Montana 
-- 
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-- 
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  INET: [EMAIL PROTECTED]

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also send the HELP command for other information (like subscribing).



RE: SQL Brain Teaser Challenge

2002-11-05 Thread Jamadagni, Rajendra
Title: RE: SQL Brain Teaser Challenge





SELECT *
 FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder) 
START WITH parentid=0 
CONNECT BY PRIOR ID = parentid


Does this work? 
Also see http://asktom.oracle.com/pls/ask/f?p=4950:8:F4950_P8_DISPLAYID:9212348049


Raj
__
Rajendra Jamadagni  MIS, ESPN Inc.
Rajendra dot Jamadagni at ESPN dot com
Any opinion expressed here is personal and doesn't reflect that of ESPN Inc. 
QOTD: Any clod can have facts, but having an opinion is an art!



*This e-mail 
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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
Nope...

(select id, parentid, nodeorder, description
from treenode
start with parentid=0 connect by prior id = parentid)
order by nodeorder;

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 5  7  0 folder 3 item 1
 8  7  1 2nd item, 3rd folder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 7  1  2 3rd subfolder
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3

-Original Message-
Sent: Tuesday, November 05, 2002 11:52 AM
To: [EMAIL PROTECTED]; Orr, Steve


Does an in-line view do the trick?
select * from
(select id, parentid, nodeorder, description
from treenode
start with parentid=0 connect by prior id = parentid)
order by nodeorder;

Jay

 [EMAIL PROTECTED] 11/05/02 12:24PM 
Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
id  number  not null 
constraint pk_treenode primary key,
parentidnumber  not null,
nodeorder   number  not null,
description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2
-

Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
Nope...

SQL l
  1  SELECT *
  2FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder)
  3  START WITH parentid=0
  4* CONNECT BY PRIOR ID = parentid
SQL /
  FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder)
  *
ERROR at line 2:
ORA-01472: cannot use CONNECT BY on view with DISTINCT, GROUP BY, etc.


I supplied the DDL/DML so you could test on your own systems.



-Original Message-
Sent: Tuesday, November 05, 2002 12:07 PM
To: '[EMAIL PROTECTED]'; Orr, Steve


SELECT * 
  FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder) 
START WITH parentid=0 
CONNECT BY PRIOR ID = parentid 
Does this work? 
Also see
http://asktom.oracle.com/pls/ask/f?p=4950:8:F4950_P8_DISPLAYID:921234804
9 
Raj 
__ 
Rajendra Jamadagni  MIS, ESPN Inc. 
Rajendra dot Jamadagni at ESPN dot com 
Any opinion expressed here is personal and doesn't reflect that of ESPN Inc.

QOTD: Any clod can have facts, but having an opinion is an art! 
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Re: ORA-02050

2002-11-05 Thread Ray Stell
On Tue, Nov 05, 2002 at 09:24:04AM -0800, Yechiel Adar wrote:
 There is another view, DBA_2PC_NEIGHBORS.
 Maybe there is some info there.

No, the transaction was rolled back completely, or perhaps
it resynced when the remote db recovered and thus covered
the tracks.  This is a good thing.

I am wondering if I can find the app that was hit.
I'm just curious, not trying to recover.  Otherwise,
flames would be coming from my hair. ;) 




 
 Yechiel Adar
 Mehish
 - Original Message -
 To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
 Sent: Tuesday, November 05, 2002 5:58 PM
 
 
  On Tue, Nov 05, 2002 at 06:33:44AM -0800, [EMAIL PROTECTED]
 wrote:
   Ray,
   I think you need to check DBA_2PC_PENDING on both the local and remote
   database to see if anything is still in there.
   That gives the osuser, terminal and host name so you may be able to get
   something from that.
   Of course if there is nothing in that table then the transaction has
 been
   rolled back already
 
 
  Right, nothing there, so I was wondering if there was a way to chase
  the transaction id to app, such as archivelog.   8.1.7.4 here.
 
  Thanks.
 
 
 
 
 
   John
  
   -Original Message-
   Sent: 05 November 2002 13:43
   To: Multiple recipients of list ORACLE-L
  
  
  
   ORA-02050: transaction 8.82.26033 rolled back, some remote DBs may be
   in-doubt
  
   The transaction seems to have completely rolled back.  My
   question is, is there any way to relate transaction id 8.82.26033
   to an application or table row or something at a higher layer?
  
   Thanks.
   ===
   Ray Stell   [EMAIL PROTECTED] (540) 231-4109 KE4TJC28^D
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   Author: Ray Stell
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Re: OCP Exams - What to study out of?

2002-11-05 Thread KENNETH JANUSZ
I agree that the SQL exam is very difficult.  I've taken it twice and missed
passing by just a couple of points.  But I will persist.

Interestingly enough when I got my APICS CPIM certification I took six exams
and passed them all on the first try.  The passing grade is 90%.

Ken Janusz, CPIM

- Original Message -
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 11:03 AM


 Eva - Everything I've heard says that all the exam questions come out of
the
 Oracle University class Student Guides. To put it bluntly, your goal is to
 pass the exam. You could study the Oracle manuals, but there is a lot of
 material to cover and what strikes you as important might not be covered
on
 the exam, and vice-versa. Essentially you are getting two levels of
 filtering. Somebody read the manuals and used that to prepare student
 guides, then a group of Oracle instructors read the student guide and
 prepared questions. I think that most of the exam guides were prepared
from
 the Oracle Student Guides and the author has at least taken the exam, so
 they have a general idea of how the test is approached. Everyone has their
 favorite exam preparation book and I consider this a worthwhile
investment,
 I bought Couchman -

http://search.barnesandnoble.com/booksearch/isbnInquiry.asp?userid=6WIANMIL0
 H

http://search.barnesandnoble.com/booksearch/isbnInquiry.asp?userid=6WIANMIL

0Hsourceid=00400731866779927243bfdate=11%2D05%2D2002+10%3A16%3A17isbn=007
 2133414

sourceid=00400731866779927243bfdate=11%2D05%2D2002+10%3A16%3A17isbn=00721
 33414
Two issues: Which exam track are you planning to take? 9i? 8i?
The advice I received is don't take the SQL exam as your first exam.
 Take the DBA exam first. The SQL exam is reported to be quite hard because
 it covers all the odd SQL functions and can really catch you unawares. As
a
 practicing DBA you should find the DBA exam pretty easy. I didn't
personally
 take that path for my own reasons, but I still consider it good advice.



 Dennis Williams
 DBA, 40%OCP
 Lifetouch, Inc.
 [EMAIL PROTECTED]

 -Original Message-
 Sent: Tuesday, November 05, 2002 6:09 AM
 To: Multiple recipients of list ORACLE-L



 Hello Everyone,

 1. I have now made the humungous decision to start studying and to write
the
 OCP exams.
 2. Do I study out of the Oracle Manuals?
 3. I do have the Sybex Study Guides, would studying these be all that is
 needed?
 4. Or do the questions come out of the Course material ( Which I have not
 attended ).
 5. I do have some of the Oreilly Insect Books - would these be any help?

 Any views, opinions etc appreciated.
 Regards
 Denham Eva
 Oracle DBA
 UNIX is basically a simple operating system, but you have to be a genius
to
 understand the simplicity.
 Dennis Ritchie.



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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
Hey Raj,

Ever paranoid of a ruse I had to see for myself that your query worked on
O9i and not O8i. I cranked up my O9i test server and confirmed it does
indeed work on 9i and not 8i. (See below.) I'm still looking for a way to
get this to work on O8i. In the meantime I'll submit a TAR but I hate it
when the solution is to upgrade to the next version.


Steve

---
SQL*Plus: Release 9.2.0.1.0 - Production on Tue Nov 5 04:58:14 2002

Copyright (c) 1982, 2002, Oracle Corporation.  All rights reserved.


Connected to:
Oracle9i Enterprise Edition Release 9.2.0.1.0 - Production
With the Partitioning, OLAP and Oracle Data Mining options
JServer Release 9.2.0.1.0 - Production



you are logged into
   RAC1.


RAC1.SYS.SQLSELECT *
  2  FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder)
  3  START WITH parentid=0
  4  CONNECT BY PRIOR ID = parentid ;

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
10  3  1 nested folder2.2.1
11 10  2 nested folder2.2.2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

11 rows selected.



-Original Message-
Sent: Tuesday, November 05, 2002 12:19 PM
To: Multiple recipients of list ORACLE-L


Steve, 
I did try it ... 
oraclei@rhea-ACPT1 sys 
SQL*Plus: Release 9.2.0.1.0 - Production on Tue Nov 5 14:18:02 2002 
Copyright (c) 1982, 2002, Oracle Corporation.  All rights reserved. 
Connected. 
SQL conn system 
Enter password: 
Connected. 
SQL SELECT * 
  FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder) 
START WITH parentid=0 
CONNECT BY PRIOR ID = parentid 
SQL / 
ID   PARENTID  NODEORDER DESCRIPTION 
-- -- --  
 1  0  0 top folder 
 9  1  0 1st subfolder 
 2  1  1 2nd subfolder 
 4  2  1 folder 2 item 1 
 3  2  2 folder 2 item 2 
 6  2  3 folder 2 item 3 
 7  1  2 3rd subfolder 
 5  7  0 folder 3 item 1 
 8  7  1 folder 3 item 2 
9 rows selected. 


Raj 
__ 
Rajendra Jamadagni  MIS, ESPN Inc. 
Rajendra dot Jamadagni at ESPN dot com 
Any opinion expressed here is personal and doesn't reflect that of ESPN Inc.

QOTD: Any clod can have facts, but having an opinion is an art! 


-Original Message- 
Sent: Tuesday, November 05, 2002 2:15 PM 
To: Jamadagni, Rajendra; '[EMAIL PROTECTED]' 


Nope... 
SQL l 
  1  SELECT * 
  2FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder) 
  3  START WITH parentid=0 
  4* CONNECT BY PRIOR ID = parentid 
SQL / 
  FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder) 
  * 
ERROR at line 2: 
ORA-01472: cannot use CONNECT BY on view with DISTINCT, GROUP BY, etc. 


I supplied the DDL/DML so you could test on your own systems. 



-Original Message- 
Sent: Tuesday, November 05, 2002 12:07 PM 
To: '[EMAIL PROTECTED]'; Orr, Steve 


SELECT * 
  FROM (SELECT * FROM treenode ORDER BY parentid, nodeorder) 
START WITH parentid=0 
CONNECT BY PRIOR ID = parentid 
Does this work? 
Also see 
http://asktom.oracle.com/pls/ask/f?p=4950:8:F4950_P8_DISPLAYID:921234804

9 
Raj 
__ 
Rajendra Jamadagni  MIS, ESPN Inc. 
Rajendra dot Jamadagni at ESPN dot com 
Any opinion expressed here is personal and doesn't reflect that of ESPN Inc.

QOTD: Any clod can have facts, but having an opinion is an art! 
-- 
Please see the official ORACLE-L FAQ: http://www.orafaq.com
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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
Well it works but your query assumes knowledge of the tree- that it will
always only have 3 levels. Consider when I add the following 2 rows:

insert into treenode values(10,3,1,'nested folder2.2.1');
insert into treenode values(11,10,2,'nested folder2.2.2');

Now it fails. SORRY I wasn't clear on this part of the rules/spec. :-( 

We should be able to add nodes and levels. We should also be able update
nodes to different parents and their children and children's children (etc)
should automatically following them around on the tree. 

Any other ideas?


Steve


-Original Message-
Sent: Tuesday, November 05, 2002 12:20 PM
To: [EMAIL PROTECTED]
Cc: Orr, Steve
Importance: High


Steve,

This works for me.

Jared



col nodelevel noprint
col parent noprint
col child noprint

select
   a.nodelevel
   , a.id id
   , a.parentid
   , a.nodeorder
   , a.description
   , decode(c.children,null,'N','Y') parent
   , decode(p.children,null,'Y','N') child
from (
   select level nodelevel, id, parentid, nodeorder, description
   from treenode
   start with parentid=0
   connect by prior id = parentid
) a,
(
   select parentid, count(*) children
   from treenode b
   group by parentid
) c,
(
   select parentid, count(*) children
   from treenode d
   group by parentid
) p
where a.id = c.parentid(+)
and a.id = p.parentid(+)
order by
   decode(parent
  -- is a parent
  , 'Y', nodelevel * id
  -- is a child
  , 'N', nodelevel * parentid + nodeorder
   )
/






Orr, Steve [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 09:24 AM
 Please respond to ORACLE-L

 
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
cc: 
Subject:SQL Brain Teaser Challenge


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
 id  number  not null 
 constraint pk_treenode 
primary key,
 parentidnumber  not null,
 nodeorder   number  not null,
 description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2
-

Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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-- 

RE: OCP Exams - What to study out of?

2002-11-05 Thread DENNIS WILLIAMS
Paula - You're losing me on the connection between SQL and GUI. I can see a
connection between command-line DBA actions and GUI DBA tools. And some GUI
report writing tools create SQL for you, but usually we DBAs are called when
it creates bad SQL code and we must straighten it out.  
 So have you passed all 5 OCP modules?



Dennis Williams 
DBA, 40%OCP 
Lifetouch, Inc. 
[EMAIL PROTECTED] 

-Original Message-
Sent: Tuesday, November 05, 2002 12:29 PM
To: Multiple recipients of list ORACLE-L



I took the SQL exam first and had no problems although I have heard others
have.  I think it has to do with how I learned - I forced myself to use SQL
before getting hooked into a GUI tool of any kind.  In the long-run that is
a good approach as you will always have that but might not always have the
GUI.

-Original Message- 
mailto:DWILLIAMS;LIFETOUCH.COM ] 
Sent: Tuesday, November 05, 2002 12:04 PM 
To: Multiple recipients of list ORACLE-L 


Eva - Everything I've heard says that all the exam questions come out of the

Oracle University class Student Guides. To put it bluntly, your goal is to 
pass the exam. You could study the Oracle manuals, but there is a lot of 
material to cover and what strikes you as important might not be covered on 
the exam, and vice-versa. Essentially you are getting two levels of 
filtering. Somebody read the manuals and used that to prepare student 
guides, then a group of Oracle instructors read the student guide and 
prepared questions. I think that most of the exam guides were prepared from 
the Oracle Student Guides and the author has at least taken the exam, so 
they have a general idea of how the test is approached. Everyone has their 
favorite exam preparation book and I consider this a worthwhile investment, 
I bought Couchman - 
http://search.barnesandnoble.com/booksearch/isbnInquiry.asp?userid=6WIANMIL0
http://search.barnesandnoble.com/booksearch/isbnInquiry.asp?userid=6WIANMIL
0  
H 

http://search.barnesandnoble.com/booksearch/isbnInquiry.asp?userid=6WIANMIL
http://search.barnesandnoble.com/booksearch/isbnInquiry.asp?userid=6WIANMIL
  
0Hsourceid=00400731866779927243bfdate=11%2D05%2D2002+10%3A16%3A17isbn=007

2133414 
sourceid=00400731866779927243bfdate=11%2D05%2D2002+10%3A16%3A17isbn=00721

33414 
   Two issues: Which exam track are you planning to take? 9i? 8i? 
   The advice I received is don't take the SQL exam as your first exam. 
Take the DBA exam first. The SQL exam is reported to be quite hard because 
it covers all the odd SQL functions and can really catch you unawares. As a 
practicing DBA you should find the DBA exam pretty easy. I didn't personally

take that path for my own reasons, but I still consider it good advice. 



Dennis Williams 
DBA, 40%OCP 
Lifetouch, Inc. 
[EMAIL PROTECTED] 

-Original Message- 
Sent: Tuesday, November 05, 2002 6:09 AM 
To: Multiple recipients of list ORACLE-L 



Hello Everyone, 

1. I have now made the humungous decision to start studying and to write the

OCP exams. 
2. Do I study out of the Oracle Manuals? 
3. I do have the Sybex Study Guides, would studying these be all that is 
needed? 
4. Or do the questions come out of the Course material ( Which I have not 
attended ). 
5. I do have some of the Oreilly Insect Books - would these be any help? 

Any views, opinions etc appreciated. 
Regards 
Denham Eva 
Oracle DBA 
UNIX is basically a simple operating system, but you have to be a genius to

understand the simplicity. 
Dennis Ritchie. 



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RE: SQL Brain Teaser Challenge

2002-11-05 Thread STEVE OLLIG
and i got this to work:
SQL select * from treenode;

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

9 rows selected.
/cheating with the order of the insert statements  ;-)

i had always thought the parentid was the way to go when representing trees
in the database, but Celko describes the Nested-Set Model of Trees
alternative in this ancient article:
http://www.dbmsmag.com/9603d06.html
makes your problem trivial.

anyway, thanks for the post Steve.  learned something today - i'm going
home.

-Original Message-
Sent: Tuesday, November 05, 2002 1:15 PM
To: Multiple recipients of list ORACLE-L


Steve,

This works for me.

Jared



col nodelevel noprint
col parent noprint
col child noprint

select
   a.nodelevel
   , a.id id
   , a.parentid
   , a.nodeorder
   , a.description
   , decode(c.children,null,'N','Y') parent
   , decode(p.children,null,'Y','N') child
from (
   select level nodelevel, id, parentid, nodeorder, description
   from treenode
   start with parentid=0
   connect by prior id = parentid
) a,
(
   select parentid, count(*) children
   from treenode b
   group by parentid
) c,
(
   select parentid, count(*) children
   from treenode d
   group by parentid
) p
where a.id = c.parentid(+)
and a.id = p.parentid(+)
order by
   decode(parent
  -- is a parent
  , 'Y', nodelevel * id
  -- is a child
  , 'N', nodelevel * parentid + nodeorder
   )
/






Orr, Steve [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 09:24 AM
 Please respond to ORACLE-L

 
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
cc: 
Subject:SQL Brain Teaser Challenge


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
 id  number  not null 
 constraint pk_treenode 
primary key,
 parentidnumber  not null,
 nodeorder   number  not null,
 description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2
-

Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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To 

RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
I was also able to confirm this works on O9i. 

-Original Message-
Sent: Tuesday, November 05, 2002 11:14 AM
To: Multiple recipients of list ORACLE-L


I get an error on 8.1.7.2. Is siblings new?

SQL l
  1  SELECT LEVEL, treenode.*
  2FROM treenode
  3   START WITH parentid=0
  4  CONNECT BY PRIOR ID = parentid
  5* ORDER SIBLINGS BY PARENTid , nodeorder
SQL /
ORDER SIBLINGS BY PARENTid , nodeorder
  *
ERROR at line 5:
ORA-00924: missing BY keyword


-Original Message-
Sent: Tuesday, November 05, 2002 11:02 AM
To: '[EMAIL PROTECTED]'; Orr, Steve


SELECT LEVEL, treenode.* 
  FROM treenode 
 START WITH parentid=0 
CONNECT BY PRIOR ID = parentid 
ORDER SIBLINGS BY PARENTid , nodeorder 
Raj 
__ 
Rajendra Jamadagni  MIS, ESPN Inc. 
Rajendra dot Jamadagni at ESPN dot com 
Any opinion expressed here is personal and doesn't reflect that of ESPN Inc.

QOTD: Any clod can have facts, but having an opinion is an art! 


-Original Message- 
Sent: Tuesday, November 05, 2002 12:24 PM 
To: Multiple recipients of list ORACLE-L 


Challenge: present SQL results hierarchically and sort the nodes. Use sort 
column without changing data. Here's the DDL/DML to start: 
create table treenode ( 
id  number  not null 
constraint pk_treenode primary key, 
parentidnumber  not null, 
nodeorder   number  not null, 
description varchar2(20)null); 
insert into treenode values(1,0,0,'top folder'); 
insert into treenode values(9,1,0,'1st subfolder'); 
insert into treenode values(7,1,2,'3rd subfolder'); 
insert into treenode values(2,1,1,'2nd subfolder'); 
insert into treenode values(8,7,1,'folder 3 item 2'); 
insert into treenode values(6,2,3,'folder 2 item 3'); 
insert into treenode values(5,7,0,'folder 3 item 1'); 
insert into treenode values(3,2,2,'folder 2 item 2'); 
insert into treenode values(4,2,1,'folder 2 item 1'); 
- 
Here's the data presented hierachically without the desired sort: 
select * from treenode 
start with parentid=0 connect by prior id = parentid; 
ID   PARENTID  NODEORDER DESCRIPTION 
-- -- --  
 1  0  0 top folder 
 9  1  0 1st subfolder 
 7  1  2 3rd subfolder 
 8  7  1 folder 3 item 2 
 5  7  0 folder 3 item 1 
 2  1  1 2nd subfolder 
 6  2  3 folder 2 item 3 
 3  2  2 folder 2 item 2 
 4  2  1 folder 2 item 1 
- 
Desired SQL statement results: 
ID   PARENTID  NODEORDER DESCRIPTION 
-- -- --  
 1  0  0 top folder 
 9  1  0 1st subfolder 
 2  1  1 2nd subfolder 
 4  2  1 folder 2 item 1 
 3  2  2 folder 2 item 2 
 6  2  3 folder 2 item 3 
 7  1  2 3rd subfolder 
 5  7  0 folder 3 item 1 
 8  7  1 folder 3 item 2 
- 
Kudos to anyone who can figure out how to do this via SQL. 


Steve Orr 
Bozeman, Montana 
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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Fink, Dan
Okay, my answer was almost correct (almost correct = wrong). Jared's answer
is right on, given the current data set.

What happens when the data is changed? Does ID have meaning or is it the
sequence in which the row was added? NODEORDER is 'sequential', but the
starting values vary within the parent node (sometimes start with 1, others
with 0).

Old college trick, instead of answering the question, you challenge the
validity of the question.

Can you use sql only or assume sql*plus is available?

-Original Message-
Sent: Tuesday, November 05, 2002 10:24 AM
To: Multiple recipients of list ORACLE-L


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
id  number  not null 
constraint pk_treenode primary key,
parentidnumber  not null,
nodeorder   number  not null,
description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2
-

Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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RE: Clone Production Server to Stand by Server on 8.1.7 on Win 2k

2002-11-05 Thread Arif Khan (GWL)
Hello Jared

If I run the database in Standby mode, once my production server fails, can
the standby database be opened for users to make changes (i.e. read an
write) and how do I accomplish that.

Also, once my production server is back on line, how do you suggest I should
synchronize the production server with he standby server database?

Thanks
Arif

-Original Message-
Sent: Tuesday, November 05, 2002 12:29 PM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
2k
Importance: High


First off, your process is doing a lot of unnecessary work.

Standby databases are available in 7.3.4.  I believe that
would be somewhat simpler than your current procedure.

I haven't tried it though, so I could be wrong.

Even with your current procedure, you don't need to copy
all of the files, most of the time anyway.

Build your standby database, put it in recovery mode, and 
just keep applying archive log files to it.

You may need to rebuild when a datafile is added to a 
tablespace, not sure.

In 8i, why not just use standby database?

Jared





Arif Khan (GWL) [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 06:53 AM
 Please respond to ORACLE-L

 
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
cc: 
Subject:Clone Production Server to Stand by Server on 8.1.7
on Win 2k


Hello

We currently have two Identical servers (identical in terms of both 
Hardware
and SW).
We run Oracle 7.3.4.0 (Workgroup server) on Win NT 4 on both the servers. 
We
call them as Production Server and Stand by server. The Stand by server is
passive in nature (i.e. does not do anything). Every night a batch process
shuts down Oracle instance on both the machines and copies over all the
files (Data, log, ctl etc) from Production Server to the Stand by server
(Drive to drive, directory to directory...)

In case the Production Server fails, we simply switch over the users (with 
a
different alias to the stand by server) and they are back in business.

Now, we are thinking of migrating to 8.1.7, however while trying to 
install
this version, one needs to specify a Global name which I believe has to be
unique on the network. So will the same process that I used to run (i.e.
copy all files over from production to stand by ) work??? I guess both my
servers will now have to have separate/unique Global Names. 

Is there any other approach that any of you can suggest??? 

TIA
Arif
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RE: Read Only Login with Source Visibility

2002-11-05 Thread Jared . Still
Depending on the tool, that may not work.

It wouldn't work with SQL Navigator, or MS Access.

Jared






Mercadante, Thomas F [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 10:59 AM
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Subject:RE: Read Only Login with Source Visibility


better yet - create a local copy of it, and place a private synonym in 
their
account to point to it.

really sneaky.


-Original Message-
Sent: Tuesday, November 05, 2002 12:24 PM
To: Multiple recipients of list ORACLE-L


Rewrite the all_source view, and I believe the all_objects view.

It's a pain in the rear, but it can be done.

It also tends to break when you upgrade.

I did this once on Oracle 7, there may be some other workaround
now on 8i.

Jared






Bill Buchan [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 04:48 AM
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[EMAIL PROTECTED]
cc: 
Subject:Read Only Login with Source Visibility


Hi all,

I am trying to create a read-only login, RO_USER which can do the 
following:

1. See all tables, views, constraints etc. in one  *specific* other 
schema, 
APP.

2. See all the source for PL/SQL objects in APP.

The first bit is easy: GRANT SELECT on tables, views to RO_USER.

Not sure about the second bit.  I have tried granting CREATE ANY PROCEDURE 


and CREATE ANY TRIGGER to RO_USER and using a DDL trigger to prevent these 


privileges actually being used to create procedures/triggers.  This works 
but does not restrict the source visibility to APP.  I have other schemas 
where I do not want RO_USER to see the source.

My other concern is granting SELECT on sequences as this means that they 
can select nextval from them and hence increment the numbers (not quite 
read-only!)

Any suggestions for fixing these problems would be much appreciated!

Thanks
- Bill.

PS. This is on 8i.


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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
Other posts should answer most of your questions. Jared's answer assumes
only three levels and doesn't work when we add levels or branches to the
tree. ID is a system generated key and has no meaning. SQL only, no SQL*Plus
stuff. NODEORDER only refers to the sort within a node because you should
not have to resequence all the data just to add a node, change the sort
within a node or move a node to another parent. 



-Original Message-
Sent: Tuesday, November 05, 2002 1:19 PM
To: Multiple recipients of list ORACLE-L


Okay, my answer was almost correct (almost correct = wrong). Jared's answer
is right on, given the current data set.

What happens when the data is changed? Does ID have meaning or is it the
sequence in which the row was added? NODEORDER is 'sequential', but the
starting values vary within the parent node (sometimes start with 1, others
with 0).

Old college trick, instead of answering the question, you challenge the
validity of the question.

Can you use sql only or assume sql*plus is available?

-Original Message-
Sent: Tuesday, November 05, 2002 10:24 AM
To: Multiple recipients of list ORACLE-L


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
id  number  not null 
constraint pk_treenode primary key,
parentidnumber  not null,
nodeorder   number  not null,
description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2
-

Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
I wondered if someone would try that. :-)

The link you provided is interesting but I don't think it fits the needs as
regards being able to add and rearrange nodes. (Maybe I got confused with
the little worms.) This is because the right column has to be twice the
number of rows and this number would need to be updated every time you added
or removed a row.


-Original Message-
Sent: Tuesday, November 05, 2002 12:45 PM
To: Multiple recipients of list ORACLE-L


and i got this to work:
SQL select * from treenode;

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

9 rows selected.
/cheating with the order of the insert statements  ;-)

i had always thought the parentid was the way to go when representing trees
in the database, but Celko describes the Nested-Set Model of Trees
alternative in this ancient article:
http://www.dbmsmag.com/9603d06.html
makes your problem trivial.

anyway, thanks for the post Steve.  learned something today - i'm going
home.

-Original Message-
Sent: Tuesday, November 05, 2002 1:15 PM
To: Multiple recipients of list ORACLE-L


Steve,

This works for me.

Jared



col nodelevel noprint
col parent noprint
col child noprint

select
   a.nodelevel
   , a.id id
   , a.parentid
   , a.nodeorder
   , a.description
   , decode(c.children,null,'N','Y') parent
   , decode(p.children,null,'Y','N') child
from (
   select level nodelevel, id, parentid, nodeorder, description
   from treenode
   start with parentid=0
   connect by prior id = parentid
) a,
(
   select parentid, count(*) children
   from treenode b
   group by parentid
) c,
(
   select parentid, count(*) children
   from treenode d
   group by parentid
) p
where a.id = c.parentid(+)
and a.id = p.parentid(+)
order by
   decode(parent
  -- is a parent
  , 'Y', nodelevel * id
  -- is a child
  , 'N', nodelevel * parentid + nodeorder
   )
/






Orr, Steve [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 09:24 AM
 Please respond to ORACLE-L

 
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
cc: 
Subject:SQL Brain Teaser Challenge


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
 id  number  not null 
 constraint pk_treenode 
primary key,
 parentidnumber  not null,
 nodeorder   number  not null,
 description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1
-
Desired SQL statement results:
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 

RE: Clone Production Server to Stand by Server on 8.1.7 on Win 2k

2002-11-05 Thread Quamrul Polash

Hi Arif,
Please read the article Note: 76373.1 on Metalink about Standby database. 

Regards,
Quamrul Polash

From: "Arif Khan (GWL)" <[EMAIL PROTECTED]>
Reply-To: [EMAIL PROTECTED] 
To: Multiple recipients of list ORACLE-L <[EMAIL PROTECTED]>
Subject: RE: Clone Production Server to Stand by Server on 8.1.7 on Win 2k 
Date: Tue, 05 Nov 2002 12:28:36 -0800 

Hello Jared 

If I run the database in Standby mode, once my production server fails, can 
the standby database be opened for users to make changes (i.e. read an 
write) and how do I accomplish that. 

Also, once my production server is back on line, how do you suggest I should 
synchronize the production server with he standby server database? 

Thanks 
Arif 

-Original Message- 
Sent: Tuesday, November 05, 2002 12:29 PM 
To: [EMAIL PROTECTED] 
Cc: [EMAIL PROTECTED] 
2k 
Importance: High 


First off, your process is doing a lot of unnecessary work. 

Standby databases are available in 7.3.4. I believe that 
would be somewhat simpler than your current procedure. 

I haven't tried it though, so I could be wrong. 

Even with your current procedure, you don't need to copy 
all of the files, most of the time anyway. 

Build your standby database, put it in recovery mode, and 
just keep applying archive log files to it. 

You may need to rebuild when a datafile is added to a 
tablespace, not sure. 

In 8i, why not just use standby database? 

Jared 





"Arif Khan (GWL)" <[EMAIL PROTECTED]>
Sent by: [EMAIL PROTECTED] 
11/05/2002 06:53 AM 
Please respond to ORACLE-L 


To: Multiple recipients of list ORACLE-L <[EMAIL PROTECTED]>
cc: 
Subject: Clone Production Server to Stand by Server on 8.1.7 
on Win 2k 


Hello 

We currently have two Identical servers (identical in terms of both 
Hardware 
and SW). 
We run Oracle 7.3.4.0 (Workgroup server) on Win NT 4 on both the servers. 
We 
call them as Production Server and Stand by server. The Stand by server is 
passive in nature (i.e. does not do anything). Every night a batch process 
shuts down Oracle instance on both the machines and copies over all the 
files (Data, log, ctl etc) from Production Server to the Stand by server 
(Drive to drive, directory to directory...) 

In case the Production Server fails, we simply switch over the users (with 
a 
different alias to the stand by server) and they are back in business. 

Now, we are thinking of migrating to 8.1.7, however while trying to 
install 
this version, one needs to specify a Global name which I believe has to be 
unique on the network. So will the same process that I used to run (i.e. 
copy all files over from production to stand by ) work??? I guess both my 
servers will now have to have separate/unique Global Names. 

Is there any other approach that any of you can suggest??? 

TIA 
Arif 
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Unlimited Internet access for only $21.95/month.  Try MSN! Click Here 
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RE: OCP Exams - What to study out of?

2002-11-05 Thread DENNIS WILLIAMS
Ken - Wow, I'm impressed. I took and passed one APICS module years and years
ago and I can testify that they aren't easy. And I took the module related
to my daily job.
   Can you share any tips on how to study for the SQL exam? I find I learn
MUCH more from someone that was forced to reassess their strategy than from
people that say oh, I took the manual home two nights and breezed through
it. 

Dennis Williams
DBA, 40%OCP
Lifetouch, Inc.
[EMAIL PROTECTED] 


-Original Message-
Sent: Tuesday, November 05, 2002 2:04 PM
To: Multiple recipients of list ORACLE-L


I agree that the SQL exam is very difficult.  I've taken it twice and missed
passing by just a couple of points.  But I will persist.

Interestingly enough when I got my APICS CPIM certification I took six exams
and passed them all on the first try.  The passing grade is 90%.

Ken Janusz, CPIM

- Original Message -
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 11:03 AM


 Eva - Everything I've heard says that all the exam questions come out of
the
 Oracle University class Student Guides. To put it bluntly, your goal is to
 pass the exam. You could study the Oracle manuals, but there is a lot of
 material to cover and what strikes you as important might not be covered
on
 the exam, and vice-versa. Essentially you are getting two levels of
 filtering. Somebody read the manuals and used that to prepare student
 guides, then a group of Oracle instructors read the student guide and
 prepared questions. I think that most of the exam guides were prepared
from
 the Oracle Student Guides and the author has at least taken the exam, so
 they have a general idea of how the test is approached. Everyone has their
 favorite exam preparation book and I consider this a worthwhile
investment,
 I bought Couchman -

http://search.barnesandnoble.com/booksearch/isbnInquiry.asp?userid=6WIANMIL0
 H

http://search.barnesandnoble.com/booksearch/isbnInquiry.asp?userid=6WIANMIL

0Hsourceid=00400731866779927243bfdate=11%2D05%2D2002+10%3A16%3A17isbn=007
 2133414

sourceid=00400731866779927243bfdate=11%2D05%2D2002+10%3A16%3A17isbn=00721
 33414
Two issues: Which exam track are you planning to take? 9i? 8i?
The advice I received is don't take the SQL exam as your first exam.
 Take the DBA exam first. The SQL exam is reported to be quite hard because
 it covers all the odd SQL functions and can really catch you unawares. As
a
 practicing DBA you should find the DBA exam pretty easy. I didn't
personally
 take that path for my own reasons, but I still consider it good advice.



 Dennis Williams
 DBA, 40%OCP
 Lifetouch, Inc.
 [EMAIL PROTECTED]

 -Original Message-
 Sent: Tuesday, November 05, 2002 6:09 AM
 To: Multiple recipients of list ORACLE-L



 Hello Everyone,

 1. I have now made the humungous decision to start studying and to write
the
 OCP exams.
 2. Do I study out of the Oracle Manuals?
 3. I do have the Sybex Study Guides, would studying these be all that is
 needed?
 4. Or do the questions come out of the Course material ( Which I have not
 attended ).
 5. I do have some of the Oreilly Insect Books - would these be any help?

 Any views, opinions etc appreciated.
 Regards
 Denham Eva
 Oracle DBA
 UNIX is basically a simple operating system, but you have to be a genius
to
 understand the simplicity.
 Dennis Ritchie.



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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Jesse, Rich
Steve,

I'm not sure why as of yet, but I had some success by creating two segmented
indexes.  One on PARENTID, NODEORDER and the second on NODEORDER, PARENTID:

ID PARENTID NODEORDER DESCRIPTION
1  00 top folder
9  10 1st subfolder
2  11 2nd subfolder
4  21 folder 2 item 1
3  22 folder 2 item 2
6  23 folder 2 item 3
7  12 3rd subfolder
5  70 folder 3 item 1
8  71 folder 3 item 2

Seems strange, though, and I don't have the time to research it.  We had the
exact same problem for our BOM structures and ended up writing a recursive
PL/SQL procedure and cursors to do it.  Not nice.

Also, I'm using CBO on 8.1.7.4 and have analyzed the tables and indexes w/o
histograms using DBMS_STATS.  BTW, the explain plan is a helluva lot better
with the indexes...  :)

HTH!  GL!

Rich


Rich Jesse   System/Database Administrator
[EMAIL PROTECTED]  Quad/Tech International, Sussex, WI USA


 -Original Message-
 From: Orr, Steve [mailto:sorr;rightnow.com]
 Sent: Tuesday, November 05, 2002 11:24 AM
 To: Multiple recipients of list ORACLE-L
 Subject: SQL Brain Teaser Challenge
 
 
 Challenge: present SQL results hierarchically and sort the 
 nodes. Use sort
 column without changing data. Here's the DDL/DML to start:
 
 create table treenode (
   id  number  not null 
   constraint pk_treenode primary key,
   parentidnumber  not null,
   nodeorder   number  not null,
   description varchar2(20)null);
 
 insert into treenode values(1,0,0,'top folder');
 insert into treenode values(9,1,0,'1st subfolder');
 insert into treenode values(7,1,2,'3rd subfolder');
 insert into treenode values(2,1,1,'2nd subfolder');
 insert into treenode values(8,7,1,'folder 3 item 2');
 insert into treenode values(6,2,3,'folder 2 item 3');
 insert into treenode values(5,7,0,'folder 3 item 1');
 insert into treenode values(3,2,2,'folder 2 item 2');
 insert into treenode values(4,2,1,'folder 2 item 1');
 -
 Here's the data presented hierachically without the desired sort:
 select * from treenode 
 start with parentid=0 connect by prior id = parentid;
 ID   PARENTID  NODEORDER DESCRIPTION
 -- -- -- 
  1  0  0 top folder
  9  1  0 1st subfolder
  7  1  2 3rd subfolder
  8  7  1 folder 3 item 2
  5  7  0 folder 3 item 1
  2  1  1 2nd subfolder
  6  2  3 folder 2 item 3
  3  2  2 folder 2 item 2
  4  2  1 folder 2 item 1
 -
 Desired SQL statement results:
 ID   PARENTID  NODEORDER DESCRIPTION
 -- -- -- 
  1  0  0 top folder
  9  1  0 1st subfolder
  2  1  1 2nd subfolder
  4  2  1 folder 2 item 1
  3  2  2 folder 2 item 2
  6  2  3 folder 2 item 3
  7  1  2 3rd subfolder
  5  7  0 folder 3 item 1
  8  7  1 folder 3 item 2
 -
 
 Kudos to anyone who can figure out how to do this via SQL.
 
 
 Steve Orr
 Bozeman, Montana
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RE: SQL Brain Teaser Challenge

2002-11-05 Thread STEVE OLLIG
actually - i think you are correct.  the Nested-Set Model of Trees as
Celko calls it would require more maintenance than the traditional parentid
column when the tree changes...  but the advantages on the query side are
interesting.

that nodeOrder column takes you on a ride.  looks to me like your stuck with
a lack of features (or a bug?) to do what you want in 8i.  you could
hardcode the levels like Jared did, or move to 9i and using Raj's
suggestion.  did you try it with a cursor?

-Original Message-
Sent: Tuesday, November 05, 2002 3:15 PM
To: Multiple recipients of list ORACLE-L


I wondered if someone would try that. :-)

The link you provided is interesting but I don't think it fits the needs as
regards being able to add and rearrange nodes. (Maybe I got confused with
the little worms.) This is because the right column has to be twice the
number of rows and this number would need to be updated every time you added
or removed a row.


-Original Message-
Sent: Tuesday, November 05, 2002 12:45 PM
To: Multiple recipients of list ORACLE-L


and i got this to work:
SQL select * from treenode;

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

9 rows selected.
/cheating with the order of the insert statements  ;-)

i had always thought the parentid was the way to go when representing trees
in the database, but Celko describes the Nested-Set Model of Trees
alternative in this ancient article:
http://www.dbmsmag.com/9603d06.html
makes your problem trivial.

anyway, thanks for the post Steve.  learned something today - i'm going
home.

-Original Message-
Sent: Tuesday, November 05, 2002 1:15 PM
To: Multiple recipients of list ORACLE-L


Steve,

This works for me.

Jared



col nodelevel noprint
col parent noprint
col child noprint

select
   a.nodelevel
   , a.id id
   , a.parentid
   , a.nodeorder
   , a.description
   , decode(c.children,null,'N','Y') parent
   , decode(p.children,null,'Y','N') child
from (
   select level nodelevel, id, parentid, nodeorder, description
   from treenode
   start with parentid=0
   connect by prior id = parentid
) a,
(
   select parentid, count(*) children
   from treenode b
   group by parentid
) c,
(
   select parentid, count(*) children
   from treenode d
   group by parentid
) p
where a.id = c.parentid(+)
and a.id = p.parentid(+)
order by
   decode(parent
  -- is a parent
  , 'Y', nodelevel * id
  -- is a child
  , 'N', nodelevel * parentid + nodeorder
   )
/






Orr, Steve [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 09:24 AM
 Please respond to ORACLE-L

 
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
cc: 
Subject:SQL Brain Teaser Challenge


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
 id  number  not null 
 constraint pk_treenode 
primary key,
 parentidnumber  not null,
 nodeorder   number  not null,
 description varchar2(20)null);

insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-
Here's the data presented hierachically without the desired sort:
select * from treenode 
start with parentid=0 connect by prior id = parentid;
ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 7  1  2 3rd subfolder
 8  7  1 folder 3 item 2
 5  7  0 folder 3 item 1
 2  1  1 2nd subfolder
 6  2  3 folder 2 item 3
 3  2  2 folder 2 item 2
 4  2  1 folder 2 item 1

RE: SQL Brain Teaser Challenge

2002-11-05 Thread Mark Richard
Steve,

I've answered this one before (not on this list) for an Oracle 8i
database...

I guess the truth is that you really can't guarantee it, but it can be
tricked with a hint.  The trick is to access the table in correct sibling
order.  Create an index on the nodeorder column and then use an index hint
in the query

SQL create index m on treenode(nodeorder);

SQL select /*+ index(t m) */ *
  2  from treenode t
  3  start with parentid=0 connect by prior id = parentid;

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

9 rows selected.

How does this work?  Well, when using connect by prior type SQL the rows
are returned in the order in which they are input, with the except being
the hierarchical sorting.  By accessing the table via the index all rows
with nodeorder 0 will be returned first, then nodeorder 1 second, etc.  The
connect-by then does it's stuff and creates the hierarchy, but by a stroke
of luck you get the result you want.

The danger with this?  Well, I always say hints are exactly what they call
themselves - hints.  Oracle could choose to use a different index, or no
index based on the query, number of rows, etc.  If Oracle doesn't obey your
hint then it won't work.  Someone commented that a subquery with an order
by isn't allowed - using the hint like this effectively does the same and
overcomes that limitation.

What do I win?

Regards,
 Mark.

PS:  Since I don't have access to 9i I haven't heard of order by
siblings...  But it sounds like it fixes the problem correctly.




   

Orr, Steve   

sorr@rightnow   To: Multiple recipients of list ORACLE-L 
[EMAIL PROTECTED]   
.comcc:   

Sent by: Subject: RE: SQL Brain Teaser Challenge   

[EMAIL PROTECTED] 

om 

   

   

06/11/2002 

06:49  

Please respond 

to ORACLE-L

   

   





Well it works but your query assumes knowledge of the tree- that it will
always only have 3 levels. Consider when I add the following 2 rows:

insert into treenode values(10,3,1,'nested folder2.2.1');
insert into treenode values(11,10,2,'nested folder2.2.2');

Now it fails. SORRY I wasn't clear on this part of the rules/spec. :-(

We should be able to add nodes and levels. We should also be able update
nodes to different parents and their children and children's children (etc)
should automatically following them around on the tree.

Any other ideas?


Steve


-Original Message-
Sent: Tuesday, November 05, 2002 12:20 PM
To: [EMAIL PROTECTED]
Cc: Orr, Steve
Importance: High


Steve,

This works for me.

Jared



col nodelevel noprint
col parent noprint
col child noprint

select
   a.nodelevel
   , a.id id
   , a.parentid
   , a.nodeorder
   , a.description
   , decode(c.children,null,'N','Y') parent
   , decode(p.children,null,'Y','N') child
from (
   select level nodelevel, id, parentid, nodeorder, description
   from treenode
   start with parentid=0
   connect by prior id = parentid
) a,
(
   select parentid, count(*) children
   from treenode 

RE: SQL Brain Teaser Challenge

2002-11-05 Thread Kevin Lange
Very good Mark.   It worked even with the new subfolders added.   I have no
idea WHY it worked ... but it did.

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
10  3  1 nested folder2.2.1
11 10  2 nested folder2.2.2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

-Original Message-
Sent: Tuesday, November 05, 2002 3:41 PM
To: Multiple recipients of list ORACLE-L


Steve,

I've answered this one before (not on this list) for an Oracle 8i
database...

I guess the truth is that you really can't guarantee it, but it can be
tricked with a hint.  The trick is to access the table in correct sibling
order.  Create an index on the nodeorder column and then use an index hint
in the query

SQL create index m on treenode(nodeorder);

SQL select /*+ index(t m) */ *
  2  from treenode t
  3  start with parentid=0 connect by prior id = parentid;

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

9 rows selected.

How does this work?  Well, when using connect by prior type SQL the rows
are returned in the order in which they are input, with the except being
the hierarchical sorting.  By accessing the table via the index all rows
with nodeorder 0 will be returned first, then nodeorder 1 second, etc.  The
connect-by then does it's stuff and creates the hierarchy, but by a stroke
of luck you get the result you want.

The danger with this?  Well, I always say hints are exactly what they call
themselves - hints.  Oracle could choose to use a different index, or no
index based on the query, number of rows, etc.  If Oracle doesn't obey your
hint then it won't work.  Someone commented that a subquery with an order
by isn't allowed - using the hint like this effectively does the same and
overcomes that limitation.

What do I win?

Regards,
 Mark.

PS:  Since I don't have access to 9i I haven't heard of order by
siblings...  But it sounds like it fixes the problem correctly.




 

Orr, Steve

sorr@rightnow   To: Multiple recipients of list
ORACLE-L [EMAIL PROTECTED]   
.comcc:

Sent by: Subject: RE: SQL Brain Teaser
Challenge   
[EMAIL PROTECTED]

om

 

 

06/11/2002

06:49

Please respond

to ORACLE-L

 

 





Well it works but your query assumes knowledge of the tree- that it will
always only have 3 levels. Consider when I add the following 2 rows:

insert into treenode values(10,3,1,'nested folder2.2.1');
insert into treenode values(11,10,2,'nested folder2.2.2');

Now it fails. SORRY I wasn't clear on this part of the rules/spec. :-(

We should be able to add nodes and levels. We should also be able update
nodes to different parents and their children and children's children (etc)
should automatically following them around on the tree.

Any other ideas?


Steve


-Original Message-
Sent: Tuesday, November 05, 2002 12:20 PM
To: [EMAIL PROTECTED]
Cc: Orr, Steve
Importance: High


Steve,

This works for me.

Jared



col nodelevel noprint
col parent noprint
col child noprint

select
   a.nodelevel
   , a.id id
   , a.parentid
   , a.nodeorder
   , a.description
   , decode(c.children,null,'N','Y') parent
   , decode(p.children,null,'Y','N') child
from (
   select level nodelevel, id, parentid, nodeorder, description
   from treenode
   start with parentid=0
   connect by prior id = parentid
) a,
(
   select parentid, count(*) children
   from treenode b
   group by parentid
) c,
(
   select parentid, count(*) children
   from treenode d
   group by parentid
) p
where a.id = c.parentid(+)
and a.id = p.parentid(+)
order by
   decode(parent
  -- is a parent
  , 'Y', nodelevel * id
  -- is a child
  , 'N', nodelevel * parentid + nodeorder
   )
/






Orr, Steve [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 

Re: Partition Question

2002-11-05 Thread Mark Richard
I agree...

What are you trying to accomplish with partitioning?  Partitioning by year
/ month / day / whatever can make it easy to truncate / archive old data.
The only trick is to create new partitions before they are required.

Another goal of partitioning may be query execution.  You might partition a
table by a certain column what is frequently stored in a where clause.
This might restrict the query to a partition rather than the entire table
and (depending on the query) could give a performance gain.

If you are lucky partitioning will achieve both, if you are unlucky
partitioning will just introduce a maintenance hassle.  Think about why you
want to partition the table and what you expect to gain by doing it.
Whatever you do, don't partition simply because you can.

Regards,
 Mark.



   
 
Don Jerman 
 
[EMAIL PROTECTED]   To: Multiple recipients of list ORACLE-L 
[EMAIL PROTECTED]   
ate.nc.uscc:  
 
Sent by:  Subject: Re: Partition Question  
 
[EMAIL PROTECTED]
 
m  
 
   
 
   
 
05/11/2002 
 
05:54  
 
Please respond 
 
to ORACLE-L
 
   
 
   
 




It depends on your reason for partitioning -- if you mean to drop a
partition in
the future (to roll off the 1999 data or whatever) then the ID range is
potentially a valid approach, as long as ID is serial.  If you just want to
put
chunks on different disk volumes, you could use the type or even a hash
partitioning scheme.  It's down to what you're trying to accomplish, and
what is
good for one partition key is probably bad or neutral for the other.


Hamid Alavi wrote:

 Hi List,

 I have a question regarding partitioning: If I want to partition a table
 which strategy is better, like do i have to use a value which from first
day
 of using this table all those partion is using or just using first
partion,
 then second etc.
 E.G:
 If I do partion tableA based on ID range 1000, so for few month the
only
 first partion of this table will be used then second partion, but if I
 partion it on Type (1,2,3,4,5) any record can be any of these type and
from
 first day all of the partions will be used.
 Just want to check with you guys which way is better for performance?
 THanks for HELP

 Hamid Alavi
 Office 818 737-0526
 Cell818 416-5095

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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
WOW. This actually works quite well!!!

All this business is for generating an HTML tree navigation object using
SQL, Perl-CGI, and javascript (from www.treeview.net) and it's critical that
everything appear in the right order or the javascript will fail miserably.
I hate to be dependent on indexes for sorts. If an index goes south or
another is added that changes the results the whole app could come crashing
down. But it's not the first time I'd be stuck with a fragile solution that
works and I don't like it anyway.


Thanks,
Steve


-Original Message-
Sent: Tuesday, November 05, 2002 2:33 PM
To: '[EMAIL PROTECTED]'
Cc: Orr, Steve


Steve,

I'm not sure why as of yet, but I had some success by creating two segmented
indexes.  One on PARENTID, NODEORDER and the second on NODEORDER, PARENTID:

ID PARENTID NODEORDER DESCRIPTION
1  00 top folder
9  10 1st subfolder
2  11 2nd subfolder
4  21 folder 2 item 1
3  22 folder 2 item 2
6  23 folder 2 item 3
7  12 3rd subfolder
5  70 folder 3 item 1
8  71 folder 3 item 2

Seems strange, though, and I don't have the time to research it.  We had the
exact same problem for our BOM structures and ended up writing a recursive
PL/SQL procedure and cursors to do it.  Not nice.

Also, I'm using CBO on 8.1.7.4 and have analyzed the tables and indexes w/o
histograms using DBMS_STATS.  BTW, the explain plan is a helluva lot better
with the indexes...  :)

HTH!  GL!

Rich


Rich Jesse   System/Database Administrator
[EMAIL PROTECTED]  Quad/Tech International, Sussex, WI USA


 -Original Message-
 From: Orr, Steve [mailto:sorr;rightnow.com]
 Sent: Tuesday, November 05, 2002 11:24 AM
 To: Multiple recipients of list ORACLE-L
 Subject: SQL Brain Teaser Challenge
 
 
 Challenge: present SQL results hierarchically and sort the 
 nodes. Use sort
 column without changing data. Here's the DDL/DML to start:
 
 create table treenode (
   id  number  not null 
   constraint pk_treenode primary key,
   parentidnumber  not null,
   nodeorder   number  not null,
   description varchar2(20)null);
 
 insert into treenode values(1,0,0,'top folder');
 insert into treenode values(9,1,0,'1st subfolder');
 insert into treenode values(7,1,2,'3rd subfolder');
 insert into treenode values(2,1,1,'2nd subfolder');
 insert into treenode values(8,7,1,'folder 3 item 2');
 insert into treenode values(6,2,3,'folder 2 item 3');
 insert into treenode values(5,7,0,'folder 3 item 1');
 insert into treenode values(3,2,2,'folder 2 item 2');
 insert into treenode values(4,2,1,'folder 2 item 1');
 -
 Here's the data presented hierachically without the desired sort:
 select * from treenode 
 start with parentid=0 connect by prior id = parentid;
 ID   PARENTID  NODEORDER DESCRIPTION
 -- -- -- 
  1  0  0 top folder
  9  1  0 1st subfolder
  7  1  2 3rd subfolder
  8  7  1 folder 3 item 2
  5  7  0 folder 3 item 1
  2  1  1 2nd subfolder
  6  2  3 folder 2 item 3
  3  2  2 folder 2 item 2
  4  2  1 folder 2 item 1
 -
 Desired SQL statement results:
 ID   PARENTID  NODEORDER DESCRIPTION
 -- -- -- 
  1  0  0 top folder
  9  1  0 1st subfolder
  2  1  1 2nd subfolder
  4  2  1 folder 2 item 1
  3  2  2 folder 2 item 2
  6  2  3 folder 2 item 3
  7  1  2 3rd subfolder
  5  7  0 folder 3 item 1
  8  7  1 folder 3 item 2
 -
 
 Kudos to anyone who can figure out how to do this via SQL.
 
 
 Steve Orr
 Bozeman, Montana
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Please see the official ORACLE-L FAQ: http://www.orafaq.com
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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
 What do I win?
This was stated in the very first post... kudos. :-)
At the moment you and Rich Jesse are tied but I'm still not very pleased
with the solution. But unless somebody comes up with something better I'll
box you up some kudos for shipping. (I afraid to ask but what are kudos
anyway?) 

Thanks.


Tentatively yours,
Steve


-Original Message-
Sent: Tuesday, November 05, 2002 2:41 PM
To: Multiple recipients of list ORACLE-L


Steve,

I've answered this one before (not on this list) for an Oracle 8i
database...

I guess the truth is that you really can't guarantee it, but it can be
tricked with a hint.  The trick is to access the table in correct sibling
order.  Create an index on the nodeorder column and then use an index hint
in the query

SQL create index m on treenode(nodeorder);

SQL select /*+ index(t m) */ *
  2  from treenode t
  3  start with parentid=0 connect by prior id = parentid;

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

9 rows selected.

How does this work?  Well, when using connect by prior type SQL the rows
are returned in the order in which they are input, with the except being
the hierarchical sorting.  By accessing the table via the index all rows
with nodeorder 0 will be returned first, then nodeorder 1 second, etc.  The
connect-by then does it's stuff and creates the hierarchy, but by a stroke
of luck you get the result you want.

The danger with this?  Well, I always say hints are exactly what they call
themselves - hints.  Oracle could choose to use a different index, or no
index based on the query, number of rows, etc.  If Oracle doesn't obey your
hint then it won't work.  Someone commented that a subquery with an order
by isn't allowed - using the hint like this effectively does the same and
overcomes that limitation.

What do I win?

Regards,
 Mark.

PS:  Since I don't have access to 9i I haven't heard of order by
siblings...  But it sounds like it fixes the problem correctly.




 

Orr, Steve

sorr@rightnow   To: Multiple recipients of list
ORACLE-L [EMAIL PROTECTED]   
.comcc:

Sent by: Subject: RE: SQL Brain Teaser
Challenge   
[EMAIL PROTECTED]

om

 

 

06/11/2002

06:49

Please respond

to ORACLE-L

 

 





Well it works but your query assumes knowledge of the tree- that it will
always only have 3 levels. Consider when I add the following 2 rows:

insert into treenode values(10,3,1,'nested folder2.2.1');
insert into treenode values(11,10,2,'nested folder2.2.2');

Now it fails. SORRY I wasn't clear on this part of the rules/spec. :-(

We should be able to add nodes and levels. We should also be able update
nodes to different parents and their children and children's children (etc)
should automatically following them around on the tree.

Any other ideas?


Steve


-Original Message-
Sent: Tuesday, November 05, 2002 12:20 PM
To: [EMAIL PROTECTED]
Cc: Orr, Steve
Importance: High


Steve,

This works for me.

Jared



col nodelevel noprint
col parent noprint
col child noprint

select
   a.nodelevel
   , a.id id
   , a.parentid
   , a.nodeorder
   , a.description
   , decode(c.children,null,'N','Y') parent
   , decode(p.children,null,'Y','N') child
from (
   select level nodelevel, id, parentid, nodeorder, description
   from treenode
   start with parentid=0
   connect by prior id = parentid
) a,
(
   select parentid, count(*) children
   from treenode b
   group by parentid
) c,
(
   select parentid, count(*) children
   from treenode d
   group by parentid
) p
where a.id = c.parentid(+)
and a.id = p.parentid(+)
order by
   decode(parent
  -- is a parent
  , 'Y', nodelevel * id
  -- is a child
  , 'N', nodelevel * parentid + nodeorder
   )
/






Orr, Steve [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
 11/05/2002 09:24 AM
 Please respond to ORACLE-L


To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
cc:
Subject:SQL Brain Teaser Challenge


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:

create table treenode (
 id  number  not 

RE: SQL Brain Teaser Challenge

2002-11-05 Thread Mark Richard
Kevin,

Like I said - as long as the table access is via the index then all rows
are returned in nodeorder (sorted globally).  The connect-by clause then
orders the result set hierarchically.  When searching for children of a
parent it will find them in correct nodeorder order.  It works for an
infinite number of depths, although I believe connect-by only works to 256
levels anyway.

The most likely problem to encounter relates to adding indexes to the id
and parentid columns - necessary if the tree of going to be much more than
(maybe) 1000 rows.  In this instance a concatenated index of
parentid,nodeorder would probably be best, although I've never really
tested that out.

As a side note:  The first person I solved this for wanted all children of
the parent sorted alphabetically - this was achieved by indexing the
equivalent of the description column - I can imagine there are several
applications of this query.



   

Kevin Lange

[EMAIL PROTECTED]   To: Multiple recipients of list ORACLE-L 
[EMAIL PROTECTED]   
om  cc:   

Sent by: Subject: RE: SQL Brain Teaser Challenge   

[EMAIL PROTECTED] 

om 

   

   

06/11/2002 

09:08  

Please respond 

to ORACLE-L

   

   





Very good Mark.   It worked even with the new subfolders added.   I have no
idea WHY it worked ... but it did.

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
10  3  1 nested folder2.2.1
11 10  2 nested folder2.2.2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

-Original Message-
Sent: Tuesday, November 05, 2002 3:41 PM
To: Multiple recipients of list ORACLE-L


Steve,

I've answered this one before (not on this list) for an Oracle 8i
database...

I guess the truth is that you really can't guarantee it, but it can be
tricked with a hint.  The trick is to access the table in correct sibling
order.  Create an index on the nodeorder column and then use an index hint
in the query

SQL create index m on treenode(nodeorder);

SQL select /*+ index(t m) */ *
  2  from treenode t
  3  start with parentid=0 connect by prior id = parentid;

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

9 rows selected.

How does this work?  Well, when using connect by prior type SQL the rows
are returned in the order in which they are input, with the except being
the hierarchical sorting.  By accessing the table via the index all rows
with nodeorder 0 will be returned first, then nodeorder 1 second, etc.  The
connect-by then does it's stuff and creates the hierarchy, but by a stroke
of luck you get the result you want.

The danger with 

RE: SQL Brain Teaser Challenge

2002-11-05 Thread Mark Richard
Steve,

I thought kudos was the codename for the latest and greatest Palm Pilot. :
-)

As I said - it's not a nice solution, it has it's limits, but it works on
Oracle 8i (and should work on other versions as well)

Regards,
 Mark.

PS:  For the sake of completeness - the parentid of node 11 should be 3
judging by the description.



   

Orr, Steve   

sorr@rightnow   To: Multiple recipients of list ORACLE-L 
[EMAIL PROTECTED]   
.comcc:   

Sent by: Subject: RE: SQL Brain Teaser Challenge   

[EMAIL PROTECTED] 

om 

   

   

06/11/2002 

09:23  

Please respond 

to ORACLE-L

   

   





 What do I win?
This was stated in the very first post... kudos. :-)
At the moment you and Rich Jesse are tied but I'm still not very pleased
with the solution. But unless somebody comes up with something better I'll
box you up some kudos for shipping. (I afraid to ask but what are kudos
anyway?)

Thanks.


Tentatively yours,
Steve


-Original Message-
Sent: Tuesday, November 05, 2002 2:41 PM
To: Multiple recipients of list ORACLE-L


Steve,

I've answered this one before (not on this list) for an Oracle 8i
database...

I guess the truth is that you really can't guarantee it, but it can be
tricked with a hint.  The trick is to access the table in correct sibling
order.  Create an index on the nodeorder column and then use an index hint
in the query

SQL create index m on treenode(nodeorder);

SQL select /*+ index(t m) */ *
  2  from treenode t
  3  start with parentid=0 connect by prior id = parentid;

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

9 rows selected.

How does this work?  Well, when using connect by prior type SQL the rows
are returned in the order in which they are input, with the except being
the hierarchical sorting.  By accessing the table via the index all rows
with nodeorder 0 will be returned first, then nodeorder 1 second, etc.  The
connect-by then does it's stuff and creates the hierarchy, but by a stroke
of luck you get the result you want.

The danger with this?  Well, I always say hints are exactly what they call
themselves - hints.  Oracle could choose to use a different index, or no
index based on the query, number of rows, etc.  If Oracle doesn't obey your
hint then it won't work.  Someone commented that a subquery with an order
by isn't allowed - using the hint like this effectively does the same and
overcomes that limitation.

What do I win?

Regards,
 Mark.

PS:  Since I don't have access to 9i I haven't heard of order by
siblings...  But it sounds like it fixes the problem correctly.






Orr, Steve

sorr@rightnow   To: Multiple recipients of
list
ORACLE-L [EMAIL PROTECTED]
.comcc:

Sent by: Subject: RE: SQL Brain Teaser
Challenge
[EMAIL PROTECTED]

om





06/11/2002

06:49

Please respond

to 

Oracle iSQL*Plus buffer overflow vulnerability (#NISR04112002)

2002-11-05 Thread Jared . Still
Alert available at:

http://metalink.oracle.com/metalink/plsql/ml2_documents.showDocument?p_database_id=NOTp_id=216775.1


- Forwarded by Jared Still/Radisys_Corporation/US on 11/05/2002 03:11 
PM -


NGSSoftware Insight Security Research [EMAIL PROTECTED]
 11/04/2002 09:48 AM

 
To: [EMAIL PROTECTED], [EMAIL PROTECTED], 
[EMAIL PROTECTED]
cc: 
Subject:Oracle iSQL*Plus buffer overflow vulnerability (#NISR04112002)


NGSSoftware Insight Security Research Advisory

Name: Oracle iSQL*Plus buffer overflow
Systems: Oracle Database 9i R1,2 on all operating systems
Severity: High Risk
Vendor URL: http://www.oracle.com/
Author: David Litchfield ([EMAIL PROTECTED])
Advisory URL: http://www.ngssoftware.com/advisories/ora-isqlplus.txt
Date: 4th November 2002
Advisory number: #NISR04112002


Description
***
Oracle iSQL*Plus is a web based application that allows users to query the
database. It is installed with Oracle 9 database server and runs on top of
apache. The iSQL*Plus module is vulnerable to a classic buffer overflow
vulnerability.


Details
***
The iSQL*Plus web application requires users to log in. After accessing 
the
default url, /isqlplus a user is presented with a log in screen. By
sending the web server an overly long user ID parameter, an internal 
buffer
is overflow on the stack and the saved return address is overwritten. This
can allow an attacker to run arbitrary code in the security context of the
web server. On most systems this will be the oracle user and on Windows
the SYSTEM user. Once the web server has been compromised attackers may
then use it as a staging platform to launch attacks against the database
server itself.

Fix Information
***
NGSSoftware alerted Oracle to this problem on the 18th of October and
Oracle, last week, issued an alert. The Oracle bug number assigned to this
issue is 2581911. Patches can be downloaded from the Oracle Metalink site
http://metalink.oracle.com/.



-- 
Please see the official ORACLE-L FAQ: http://www.orafaq.com
-- 
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Oracle9iAS Web Cache Denial of Service (a102802-1)

2002-11-05 Thread Jared . Still
FYI
- Forwarded by Jared Still/Radisys_Corporation/US on 11/05/2002 03:12 
PM -


@stake advisories advisories
 10/28/2002 11:05 AM

 
To: [EMAIL PROTECTED]
cc: 
Subject:Oracle9iAS Web Cache Denial of Service (a102802-1)


-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1


   @stake, Inc.
 www.atstake.com

Security Advisory


Advisory Name: Oracle9iAS Web Cache Denial of Service
 Release Date: 10-28-2002
  Application: Oracle9iAS Web Cache 9.0.2.0.0
 Platform: Windows NT/2000/XP
 Severity: Remote anonymous DoS
   Author: Andreas Junestam ([EMAIL PROTECTED])
Vendor Status: Oracle has released a bulletin
CVE Candidate: CAN-2002-0386 
Reference: www.atstake.com/research/advisories/2002/a102802-1.txt


Overview:

Oracle Web Cache is a part of the Oracle Application Server suite. The
Web Cache server is designed to be implemented in front of the Oracle
Web server and act as a caching reverse proxy server.

There exists two different denial of service scenarios, which will cause
the Web Cache service to fail. The denial of service conditions can be
exploited by simple HTTP requests to the Web Cache service.


Detailed Description:

There exists two different denial of service situations in Oracle Web
Cache 9.0.2.0.0. The first one is triggered by issuing a HTTP GET
request containing at least one dot-dot-slash contained in the URI:

GET /../ HTTP/1.0
Host: whatever
[CRLF]
[CRLF]

The second denial of service is triggered by issuing an malformed GET
request:

GET / HTTP/1.0
Host: whatever
Transfer-Encoding: chunked
[CRLF]
[CRLF]

Both will create an exception and the service will fail.


Vendor Response:

Vendor was first contacted by @stake: 08-28-2002.
Vendor released a bulletin: 10-04-2002

Oracle has released a bulletin describing a solution to this issue.


Recommendation:

Follow the vendor's instructions detailed in the security bulletin for
this issue. 

- - From the Oracle bulletin:

  Customers should follow best security practices for protecting the
  administration process from unauthorized users and requests. As such,
  Oracle strongly encourages customers to take both of the following
  protective measures:
  1. Use firewall techniques to restrict access to the Web Cache
 administration port.
  2. Use the Secure Subnets feature of the Web Cache Manager tool to
 provide access only to administrators connecting from a list of
 permitted IP addresses or subnets.
  The potential security vulnerability is being tracked internally at
  Oracle and will be fixed by default in the 9.0.4 release of Oracle9i
  Application Server.

  For more information, see:
  http://otn.oracle.com/deploy/security/pdf/2002alert43rev1.pdf


Common Vulnerabilities and Exposures (CVE) Information:

CAN-2002-0386 Oracle9iAS Web Cache Denial of Service


@stake Vulnerability Reporting Policy:
http://www.atstake.com/research/policy/

@stake Advisory Archive:
http://www.atstake.com/research/advisories/

PGP Key:
http://www.atstake.com/research/pgp_key.asc

Copyright 2002 @stake, Inc. All rights reserved.

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Re: SQL Brain Teaser Challenge

2002-11-05 Thread Stephane Faroult
Orr, Steve wrote:
 
  What do I win?
 This was stated in the very first post... kudos. :-)
 At the moment you and Rich Jesse are tied but I'm still not very pleased
 with the solution. But unless somebody comes up with something better I'll
 box you up some kudos for shipping. (I afraid to ask but what are kudos
 anyway?)
 
 Thanks.
 
 Tentatively yours,
 Steve
 

Steve,

   I have a solution which doesn't rely on hints, but I am not very
satisfied with it either. It relies on a function, and performance will
be likely to be dismal if your tree grows big.
Here is the function :
create or replace function tree_rank(p_id in number)
return number
is
  n_rank   number;
begin
  select sum(nodeorder * power(10, -1 * level))
  into n_rank
  from treenode
  where id in (select id from treenode
   connect by id = prior parentid
   start with id = p_id)
  connect by parentid = prior id
  start with id = 1;
  return n_rank;
end;
/

(double 'CONNECT BY', ouch). Note that if you expect more than 10 items
per level, you should use somthing bigger than 10 in the power function.

However :

SQL select * from treenode
  2  order by tree_rank(id);

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

9 rows selected.

-- 
Regards,

Stephane Faroult
Oriole Software
-- 
Please see the official ORACLE-L FAQ: http://www.orafaq.com
-- 
Author: Stephane Faroult
  INET: [EMAIL PROTECTED]

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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Orr, Steve
 I have a solution which doesn't rely on hints, but I am not very
 satisfied with it either.

Innovative nonetheless. Another cool way to skin this cat.
Thanks!
Steve


-Original Message-
Sent: Tuesday, November 05, 2002 5:14 PM
To: Multiple recipients of list ORACLE-L
Importance: High


Orr, Steve wrote:
 
  What do I win?
 This was stated in the very first post... kudos. :-)
 At the moment you and Rich Jesse are tied but I'm still not very pleased
 with the solution. But unless somebody comes up with something better I'll
 box you up some kudos for shipping. (I afraid to ask but what are kudos
 anyway?)
 
 Thanks.
 
 Tentatively yours,
 Steve
 

Steve,

   I have a solution which doesn't rely on hints, but I am not very
satisfied with it either. It relies on a function, and performance will
be likely to be dismal if your tree grows big.
Here is the function :
create or replace function tree_rank(p_id in number)
return number
is
  n_rank   number;
begin
  select sum(nodeorder * power(10, -1 * level))
  into n_rank
  from treenode
  where id in (select id from treenode
   connect by id = prior parentid
   start with id = p_id)
  connect by parentid = prior id
  start with id = 1;
  return n_rank;
end;
/

(double 'CONNECT BY', ouch). Note that if you expect more than 10 items
per level, you should use somthing bigger than 10 in the power function.

However :

SQL select * from treenode
  2  order by tree_rank(id);

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

9 rows selected.

-- 
Regards,

Stephane Faroult
Oriole Software
-- 
Please see the official ORACLE-L FAQ: http://www.orafaq.com
-- 
Author: Stephane Faroult
  INET: [EMAIL PROTECTED]

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Re: suggestion w/c platforms to choose from...

2002-11-05 Thread grace
database will be used for oltp and  data mart
# of users -- 200
very critical, since order taking , purchase order and accounting  module,hr
will run on it..

- Original Message -
To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
Sent: Tuesday, November 05, 2002 9:03 PM


 As a rule buy the biggest, meanest, fault tolerance, with gigabytes of
 memory and terabytes of disk storage that you can buy.

 If you will provide more data about:

 1) The size of the database
 2) How many users
 3) How critical is the system
 4) The use of the system - data warehouse, OLTP etc

 then you will probably get a more specified answer.

 Yechiel Adar
 Mehish
 - Original Message -
 To: Multiple recipients of list ORACLE-L [EMAIL PROTECTED]
 Sent: Tuesday, November 05, 2002 2:13 PM


  hi,
   can anyone suggestion w/c platform should i used to run oracle? wat are
 the
  things to consider in choosing platform?
 
  thanks
 
  Best regards,
  Grace Lim
  MIS Department
  Suy Sing Comm'l Corp.
  T- (632)-2474134
  F- (632)-2474160
 
  --
  Please see the official ORACLE-L FAQ: http://www.orafaq.com
  --
  Author: grace
INET: [EMAIL PROTECTED]
 
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 --
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 --
 Author: Yechiel Adar
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RE: SQL Brain Teaser Challenge

2002-11-05 Thread Karniotis, Stephen
Steve:

   Unfortunately, I didn't have time to experiment with this Teaser, but I
was glad (real glad) to see that several great minds made the attempt.  I
could use all of you to help teach my students how to use their minds to
solve problems like this.  Ok.  Now back to work folks.  See you at
OracleWorld

Thank You

Stephen P. Karniotis
Product Architect
Compuware Corporation
Direct: (248) 865-4350
Mobile: (248) 408-2918
Email:  [EMAIL PROTECTED]
Web:www.compuware.com

 -Original Message-
Sent:   Tuesday, November 05, 2002 7:43 PM
To: Multiple recipients of list ORACLE-L
Subject:RE: SQL Brain Teaser Challenge

 I have a solution which doesn't rely on hints, but I am not very
 satisfied with it either.

Innovative nonetheless. Another cool way to skin this cat.
Thanks!
Steve


-Original Message-
Sent: Tuesday, November 05, 2002 5:14 PM
To: Multiple recipients of list ORACLE-L
Importance: High


Orr, Steve wrote:
 
  What do I win?
 This was stated in the very first post... kudos. :-)
 At the moment you and Rich Jesse are tied but I'm still not very pleased
 with the solution. But unless somebody comes up with something better I'll
 box you up some kudos for shipping. (I afraid to ask but what are kudos
 anyway?)
 
 Thanks.
 
 Tentatively yours,
 Steve
 

Steve,

   I have a solution which doesn't rely on hints, but I am not very
satisfied with it either. It relies on a function, and performance will
be likely to be dismal if your tree grows big.
Here is the function :
create or replace function tree_rank(p_id in number)
return number
is
  n_rank   number;
begin
  select sum(nodeorder * power(10, -1 * level))
  into n_rank
  from treenode
  where id in (select id from treenode
   connect by id = prior parentid
   start with id = p_id)
  connect by parentid = prior id
  start with id = 1;
  return n_rank;
end;
/

(double 'CONNECT BY', ouch). Note that if you expect more than 10 items
per level, you should use somthing bigger than 10 in the power function.

However :

SQL select * from treenode
  2  order by tree_rank(id);

ID   PARENTID  NODEORDER DESCRIPTION
-- -- -- 
 1  0  0 top folder
 9  1  0 1st subfolder
 2  1  1 2nd subfolder
 4  2  1 folder 2 item 1
 3  2  2 folder 2 item 2
 6  2  3 folder 2 item 3
 7  1  2 3rd subfolder
 5  7  0 folder 3 item 1
 8  7  1 folder 3 item 2

9 rows selected.

-- 
Regards,

Stephane Faroult
Oriole Software
-- 
Please see the official ORACLE-L FAQ: http://www.orafaq.com
-- 
Author: Stephane Faroult
  INET: [EMAIL PROTECTED]

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  INET: [EMAIL PROTECTED]

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The contents of this e-mail are intended for the named addressee only. It
contains information that may be confidential. Unless you are the named
addressee or an authorized designee, you may not copy or use it, or disclose
it to anyone else. If you received it in error please notify us immediately
and then destroy it. 

-- 
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/opt/oracle/product/jre/1.1.8/bin/jre

2002-11-05 Thread Sinardy Xing
Hi all,

Can I set this file to r-xr-xr-x ?

rwxrwxrwx   /opt/oracle/product/jre/1.1.8/bin/jre


Thanks
Sinardy
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RE: OCP Exams - What to study out of?

2002-11-05 Thread Denham Eva
Title: OCP Exams - What to study out of?



Thanks 
to everyone for their input. I am considering the 8i track.
a. 
Because I work mainly with the 8i DB.
b. I 
hear that for the 9i a attendance of a course is a 
prerequisite.

Regards
Denham

  -Original Message-From: Denham Eva 
  [mailto:[EMAIL PROTECTED]]Sent: Tuesday, November 05, 2002 2:09 
  PMTo: Multiple recipients of list ORACLE-LSubject: OCP 
  Exams - What to study out of?
  Hello Everyone, 
  1. I have now made the humungous decision to start 
  studying and to write the OCP exams. 2. Do 
  I study out of the Oracle Manuals? 3. I do 
  have the Sybex Study Guides, would studying these be all that is 
  needed? 4. Or do the questions come out of 
  the Course material ( Which I have not attended ). 5. I do have some of the Oreilly "Insect" Books - would these be any 
  help? 
  Any views, opinions etc appreciated. 
  Regards Denham 
  Eva Oracle DBA 
  "UNIX is basically a simple 
  operating system, but you have to be a genius to understand the 
  simplicity." Dennis 
  Ritchie. 
  
  
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of the individual sender, except where the message states otherwise and the 
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