RE: MYSQL 4.0

2002-06-27 Thread G r e g L a w r i e 

Jeremy,

Would you mind sharing how many machines 'half your servers' might be, what
type of work are they doing and how long have they been up for / how stable
would you consider them to be?

Thanks,

Greg


-Original Message-
From: Jeremy Zawodny [mailto:[EMAIL PROTECTED]]
Sent: Thursday, 27 June 2002 16:23
To: Arul
Cc: MySQL
Subject: Re: MYSQL 4.0


On Wed, Jun 26, 2002 at 08:15:11PM +0530, Arul wrote:
 Hi

 Any idea when MySQL 4.0 stable release will be made..
 As of now only Alpha is released..

Rumor has it that 4.0.2 might be beta.

 Also any sites which run on 4.0 Alpha

Yes.  Half my servers have been moved to what will be 4.0.2.
--
Jeremy D. Zawodny, [EMAIL PROTECTED]
Technical Yahoo - Yahoo Finance
Desk: (408) 349-7878   Fax: (408) 349-5454   Cell: (408) 685-5936

MySQL 3.23.51: up 28 days, processed 612,154,047 queries (250/sec. avg)

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RE: stupid error, please kick me (and send me a solution)

2002-06-03 Thread G r e g L a w r i e 

It would also appear you are missing a closing ) at the end of the
'$query=...' line. You have two opening backets and only one closing.

Greg

-Original Message-
From: Bruce Lewis [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, 4 June 2002 7:50
To: [EMAIL PROTECTED]; mysql; php-general
Subject: Re: stupid error, please kick me (and send me a solution)


Your missing your closing } at the end of the else statement.


- Original Message -
From: Jule Slootbeek [EMAIL PROTECTED]
To: mysql [EMAIL PROTECTED]; php-general
[EMAIL PROTECTED]
Sent: Monday, June 03, 2002 5:09 PM
Subject: stupid error, please kick me (and send me a solution)


 Hey guys,
 i'm getting this error with the following sql script using php:
 --error--
 Warning: Supplied argument is not a valid MySQL-Link resource in
 /var/www/phpquiz/register_user.php on line 12
 --error--

 --script--
 $link_glob = mysql_connect('$host_glob', '$un_glob', '$pw_glob');
 $query = INSERT INTO user values('0', '$fname', '$lname', '$email',
 '$username', PASSWORD('$password');
 $result = mysql_db_query('$db_glob', '$query', $link_glob');
 if (!$result) {
 echo font size=+1Your Information could not be entered into the
database,
 Please contact the a
 href=mailto:$webmasterwebmaster/a./fontbrbr . mysql_errno() .
 :  . mysql_error() . brbr;
 } else {
 echo font size=+1Your Information has successfully been entered into
the
 database!/fontbr;

 --
 Jule Slootbeek
 [EMAIL PROTECTED]

 http://blindtheory.cjb.net



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RE: Stupid - I just can't get it (NOT IN or NOT EXISTS)

2002-05-13 Thread G r e g L a w r i e 

I think you need to read the page you have refered to again.

It specifically says that sub-selects are still scheduled for implementation
and the examples given are ways of rewriting subselects to achieve the same
result using alternative methods i.e. using joins. The second example on
their page does what you are wanting.




-Original Message-
From: John Doe [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, 14 May 2002 7:27
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: Stupid - I just can't get it (NOT IN or NOT EXISTS)


Well that is really dumb because on page:
www.mysql.com/doc/A/N/ANSI_diff_Sub-selects.html
it gives me a manual page oh the syntax of how to use it.  However I think
you are right and the manual is wrong.  (congrats)
J


From: Nick Stuart [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
CC: [EMAIL PROTECTED]
Subject: Re: Stupid - I just can't get it (NOT IN or NOT EXISTS)
Date: Mon, 13 May 2002 17:45:59 -0400 (EDT)

Simple, MySQL doesn't support sub selects. There is a way around this
using JOINS but don't
know those well enough to describe...anyone else want to give it a shot?

-Nick

 
Simle
query request: I want to display the ID for everyone in a table
  who is  NOT in a second

table.
 
  so this is what I have come up with
 
  SELECT student.SID,

student.firstName
  FROM student
  WHERE student.SID NOT IN
  (SELECT

msoccerroster.SID
   FROM msoccerroster)
 
 
  It comes up with an error 1064: SQL

syntax near '(SELECT
  msoccerroster.SID  FROM msoccerroster)'
 
  please help
 

J
 
 
 
_
 
Chat with
friends online, try MSN Messenger: http://messenger.msn.com
 
 
  -
 
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posting, please check:
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RE: ASP and MySQL

2002-03-27 Thread G r e g L a w r i e 

Loretta,

Your syntax appears to be correct. Below is an example of what we do (with
some bits cut out) and it works.

strSQL = INSERT INTO tblTempConsolidation SET   _
customerId =   intCustomerId  , 
 ...
strSQL = strSQL  orderPrice = 0,   _
orderQty =   intQty  ,   _
orderSource = 3,   _
245A = '  
Replace(strTitle,','')  ',   _
100A = '  
Replace(strAuthor,','')  ', 
 ...
objConn.Execute strSQL

If you want to forward a little more code directly to me, I am more than
happy to have a look.

Regards,

Greg


-Original Message-
From: Loretta [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, 27 March 2002 16:24
To: [EMAIL PROTECTED]
Subject: ASP and MySQL


I am having a problem inserting names with apostrophes into the MySQL
database.  I am using ASP as my front end with standard SQL statements to
add, edit and delete records.  I read the section in the manual on escape
characters and have played around with many different variations trying to
get it to work with absolutely no luck.

The statement I am currently using looks like this:
lname = Replace(Request.Form(lname),','')

I have tried using the backslash character and several variations of
different numbers of apostrophes.

I would appreciate any help as this is the last problem to be solved before
the database is functional.  The rest of the debugging has been completed.

Thank-you in advance,
Loretta


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RE: Is The MySQL Mailing List On Crack?

2002-02-13 Thread G r e g L a w r i e 

I'm getting it and several others have as well !!!


 -Original Message-
 From: Jeremy Zawodny [mailto:[EMAIL PROTECTED]]
 Sent: Wednesday, 13 February 2002 11:45
 To: [EMAIL PROTECTED]
 Subject: Is The MySQL Mailing List On Crack?


 I'm getting duplicate messages separated by several hours.  Am I alone
 in this, or are others seeing it too?

 Jeremy
 --
 Jeremy D. Zawodny, [EMAIL PROTECTED]
 Technical Yahoo - Yahoo Finance
 Desk: (408) 349-7878   Fax: (408) 349-5454   Cell: (408) 685-5936

 MySQL 3.23.47-max: up 5 days, processed 189,890,555 queries (402/sec. avg)

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RE: running MySQL on WinXP

2001-12-09 Thread G r e g L a w r i e 

Haven't tried it myself, but can't see why it shouldn't.

How about you try telling us the process you have been through so far and
any messages / errors that you may be getting.


 -Original Message-
 From: Curt Kliewer [mailto:[EMAIL PROTECTED]]
 Sent: Monday, 10 December 2001 5:16
 To: MySQL List
 Subject: running MySQL on WinXP


 Hi.  I am new to PHP/MySQL and this list.

 Can I run MySQL on WinXP?  I haven't been able to get it to work
 so far.  Any help would be appreciated.
 --
 Best regards,
 Curt Kliewer
 **
 **
 Sol Interactive Development
 Applications Programmer/Analyst
 Internet Solutions/Server-Side Programmer
 **
 **
 Contact Information:
 ICQ:1835597
 MSN Messenger:   [EMAIL PROTECTED]
 Email:[EMAIL PROTECTED]
 Website:   http://www.sol-interactive.com
 **
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RE: Viewing table data in a webpage using ASP

2001-12-03 Thread G r e g L a w r i e 

Nick,

If you are able to add records, then your connection string is correct, at
least for viewing the table.

Need to see the code where you actually loop through and display each
record.

Greg



 -Original Message-
 From: Nick Boldt [mailto:[EMAIL PROTECTED]]
 Sent: Tuesday, 4 December 2001 7:49
 To: [EMAIL PROTECTED]
 Subject: RE: Viewing table data in a webpage using ASP


 Here is the connection code I use to view record information from a
 table.  I'm using Dreamweaver Ultradev 4 to help with some of the basic
 coding.  I am using MyODBC.

 %
 MM_TrackoftheWolf_STRING = dsn=DSN Name;uid=User Id;pwd=Password;

 set ship_address = Server.CreateObject(ADODB.Recordset)
 ship_address.ActiveConnection = MM_TrackoftheWolf_STRING
 ship_address.Source = SELECT * FROM customershipping
 ship_address.CursorType = 0
 ship_address.CursorLocation = 1
 ship_address.LockType = 3
 ship_address.Open()
 ship_address_numRows = 0
 %
 %
 Dim Repeat1__numRows
 Repeat1__numRows = -1
 Dim Repeat1__index
 Repeat1__index = 0
 ship_address_numRows = ship_address_numRows + Repeat1__numRows
 %

 Nick Boldt
 Computer Network Administrator
 Track of the Wolf
 Elk River, MN 55330
 Tel: 763-633-2500  Fax: 763-633-2550
 [EMAIL PROTECTED]

 -Original Message-
 From: Christopher R. Jones [mailto:[EMAIL PROTECTED]]
 Sent: Monday, December 03, 2001 3:07 PM
 To: Nick Boldt
 Subject: Re: Viewing table data in a webpage using ASP

 If you have MyODBC installed and working you should be able to view the
 records.  I have a number of WEB pages that display MySQL data using
 ASP/ADO. You should send an example of you ASP script to the group to
 help
 clarify your problem.



 Hi,
 
 I'm trying to view the records of a table in on a webpage, but nothing
 is showing up.  I can add records to the table from the page but cannot
 view, update or delete them.
 
 I'm using Windows 2000 and IIS 5.0
 
 I'm new to MySql and would greatly appreciate any help anyone can give
 me.
 
 Nick Boldt
 Computer Network Administrator
 Track of the Wolf
 Elk River, MN 55330
 Tel: 763-633-2500  Fax: 763-633-2550
 
 
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 Christopher R. Jones, P.Eng.
 14 Oneida Avenue
 Toronto, Ontario M5J 2E3
 Tel. 416 203-7465
 Fax. 416 203-8249
 Email [EMAIL PROTECTED]






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RE: MySql on redhat 6.x and above

2001-11-24 Thread G r e g L a w r i e 

Howdy,

I'm running MySQL 3.23.43 on Redhat 7.1 and an older version of MySQL
3.22.something on Redhat 6.2. Download and install from the RPM's and it
works like a charm. I have never had a problem.

One day when it is working a little harder, I shall look at recompiling to
optimise, but so far, i does it's job.

Greg Lawrie

M e l b e c
PO Box 649
Modbury  South Australia  5092

--
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 -Original Message-
 From: Gordan Bobic [mailto:[EMAIL PROTECTED]]
 Sent: Saturday, 24 November 2001 0:34
 To: [EMAIL PROTECTED]
 Subject: Re: MySql on redhat 6.x and above


 On Friday 23 Nov 2001 11:27, you wrote:
  Hi,
  I have some doubts regarding MySql on Linux
  environment.
  I'd like to install MySql on linux -redhat 6.x and
  above.
  Could you please guide me how to proceed.
  I did not find a specific document on redhat 6.x and
  above in the documentations provided in the site.

 Try with the RPMs. You may want to recompile it yourself to get the
 optimizations right for your application, though.

 I'm running it on RH7.

 Regards.

 Gordan

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SELECT COUNT(*) question ...

2001-08-20 Thread G r e g L a w r i e 

Howdy all,

I have two tables as described further down below ...

I use the following query to get the data I need ...

sql = SELECT tblTempConsolidation.*, tblCustomer.customerLibrary FROM
tblCustomer, tblTempConsolidation WHERE tblCustomer.customerId =
tblTempConsolidation.customerId AND tblTempConsolidation.customerId = 
Cint(intCustomerId)  ;

(I'm using VBScript)

I need to know how many records this SELECT returns to me. I am currently
doing this with a seperate SELECT COUNT(*) statement and it works fine,
however it would be nice to do the whole lot in one. The manual tells me I
can not mix GROUP and no GROUP columns without having a GROUP BY. From my
rather limited SQL knowledge, I suspect that I can not do this with the data
that I have..

Any suggestions? TIA.


#
# Table structure for table 'tblCustomer'
#

CREATE TABLE tblCustomer (
  customerId int(11) NOT NULL DEFAULT '0' auto_increment,
  customerName char(10) NOT NULL DEFAULT '' ,
  customerPW char(10) ,
  customerLibrary char(50) ,
  customerEmail char(50) ,
  interfaceTypeId int(11) NOT NULL DEFAULT '0' ,
  customerUSMARC tinyint(4) ,
  PRIMARY KEY (customerId),
  KEY customerName (customerName),
  KEY interfaceTypeId (interfaceTypeId)
);

#
# Table structure for table 'tblTempConsolidation'
#

CREATE TABLE tblTempConsolidation (
  sequence int(11) NOT NULL DEFAULT '0' auto_increment,
  customerId int(11) NOT NULL DEFAULT '0' ,
  020A char(10) NOT NULL DEFAULT '' ,
  orderPrice decimal(20,4) NOT NULL DEFAULT '0.' ,
  orderQty smallint(6) NOT NULL DEFAULT '0' ,
  orderSource tinyint(3) unsigned NOT NULL DEFAULT '0' ,
  245A char(50) ,
  100A char(50) ,
  260B char(50) NOT NULL DEFAULT '' ,
  GroupID int(11) NOT NULL DEFAULT '0' ,
  PRIMARY KEY (customerId,020A),
  KEY sequence (sequence),
  KEY customerId (customerId),
  KEY 020A (020A),
  KEY GroupID (GroupID)
);


Greg Lawrie

C i m t e c h
51 Byron Place
Adelaide  South Australia  5000
Phone +618 8212 8799  Fax +618 8212 8766
Email [EMAIL PROTECTED] mailto:[EMAIL PROTECTED]

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RE: Locked out of mySQL.

2001-01-27 Thread G r e g L a w r i e 

In the manual 

  First off, have you tried the following ...

6.15 Causes of Access denied Errors
If you encounter Access denied errors when you try to connect to the MySQL
server, the list below indicates some courses of action you can take to
correct the problem:

After installing MySQL, did you run the mysql_install_db script to set up
the initial grant table contents? If not, do so. See section 6.12 Setting Up
the Initial MySQL Privileges. Test the initial privileges by executing this
command:
shell mysql -u root test

The server should let you connect without error. You should also make sure
you have a file `user.MYD' in the MySQL database directory. Ordinarily, this
is `PATH/var/mysql/user.MYD', where PATH is the pathname to the MySQL
installation root.
After a fresh installation, you should connect to the server and set up your
users and their access permissions:
shell mysql -u root mysql

The server should let you connect because the MySQL root user has no
password initially. That is also a security risk, so setting the root
password is something you should do while you're setting up your other MySQL
users. If you try to connect as root and get this error:
Access denied for user: '@unknown' to database mysql

this means that you don't have an entry in the user table with a User column
value of 'root' and that mysqld cannot resolve the hostname for your client.
In this case, you must restart the server with the --skip-grant-tables
option and edit your `/etc/hosts' or `\windows\hosts' file to add an entry
for your host.


  If so, then do this ...

20.11 How to reset a forgotten password.
If you have forgotten the root user password for MySQL, you can restore it
with the following procedure.

Take down the mysqld server by sending a kill (not kill -9) to the mysqld
server. The pid is stored in a .pid file which is normally in the MySQL
database directory:
kill `cat /mysql-data-directory/hostname.pid`

You must be either the Unix root user or the same user the server runs as to
do this.
Restart mysqld with the --skip-grant-tables option.
Connect to the mysqld server with mysql -h hostname mysql and change the
password with a GRANT command. See section 7.34 GRANT and REVOKE syntax. You
can also do this with mysqladmin -h hostname -u user password 'new password'
Load the privilege tables with: mysqladmin -h hostname flush-privileges or
with the SQL command FLUSH PRIVILEGES.
Note that after you started mysqld with --skip-grant-tables, any usage of
GRANT commands will give you an Unknown command error until you have
executed FLUSH PRIVILEGES.




 -Original Message-
 From: AJ Hawks [mailto:[EMAIL PROTECTED]]
 Sent: Saturday, 27 January 2001 19:01
 To: mySQL
 Subject: Locked out of mySQL.


 Hello,
   I'm trying to learn mysql  PHP. I recently set up the
 mysql server (with
 the file: mysql-3.23.32-pc-linux-gnu-i686.tar.gz)

 My problem is this: I either forgot the password, or neglected to
 set up any
 user accounts.

 The documentation says to do this:
 ./bin/mysqladmin -u root -p password 'new-password'
 ./bin/mysqladmin -u root -h twistedlogic -p password 'new-password'

 which returns the following:

 Enter password:
 ./bin/mysqladmin: connect to server at 'localhost' failed
 error: 'Access denied for user: 'root@localhost' (Using password: YES)'

 I also tried renaming the mysql directory (in /usr/local/) and
 re-installing
 it. no. maybe I did something wrong there, because that seems to me it
 should have worked, being a binary package... and the mysql table
 being held
 in mysql/data/mysql.

 anyway, if anyone knows how I can get back in tell me! :)


 thanks,
 AJ.


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RE: rank newbie looking for SQL forum

2001-01-24 Thread G r e g L a w r i e 

I haven't participated, but the site at

www.sqlteam.com

looks useful and does have a number of forums there.

HTH,

G.


 -Original Message-
 From: Jason Witherspoon [mailto:[EMAIL PROTECTED]]
 Sent: Thursday, 25 January 2001 8:14
 To: [EMAIL PROTECTED]
 Subject: rank newbie looking for SQL forum
 
 
 Hi--
 
 As stated above, I'm a raw beginner looking for some (probably fairly 
 simple) SQL conceptual help.  Having lurked for a while, I can see that 
 since this isn't MySQL-specific, there's probably a better forum 
 in which I 
 can attempt to iron out my confusion.
 
 Can anyone point me to a mailing list/usenet group where I can 
 ask lots of 
 dumb newbie SQL questions w/o making a major pest of myself?
 
 I'll be back here in a short while, when I try to bang my head against a 
 Red Hat MySQL install again.
 
 Thanks!
 
 
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RE: ORDER BY command

2001-01-16 Thread G r e g L a w r i e 

Hmmm, you seem to have things a little backwards as ASCENDING should be the
default order.

However, as per the manual, the following will do it 

SELECT * FROM tbl_name ORDER BY col_name ASC

To sort in the opposite order 

SELECT * FROM tbl_name ORDER BY col_name DESC

Regards,

Greg


 -Original Message-
 From: Robert [mailto:[EMAIL PROTECTED]]
 Sent: Wednesday, 17 January 2001 7:10
 To: [EMAIL PROTECTED]
 Subject: ORDER BY command


 Hello. I am trying to get the results from a query to be ORDERED by
 ASCENDING instead of the default DESCENDING. Is this possible, and if so
 how?

 Robert


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