RE: Count syntax

2007-09-28 Thread Beauford
Thanks - it works, but what does the 1 and 0 do in this -
SUM(IF(supportertype = 'L', 1, 0))

 -Original Message-
 From: Baron Schwartz [mailto:[EMAIL PROTECTED] 
 Sent: September 28, 2007 1:00 PM
 To: Beauford
 Cc: mysql@lists.mysql.com
 Subject: Re: Count syntax
 
 Beauford wrote:
  Hi,
  
  I have the following line of code and I keep getting wrong 
 results from it.
  Can someone let me know what I'm doing wrong here. I just 
 can't quite 
  figure out the syntax that I need.
  
  select count(*) as numrows, count(supportertype) as leadcar from 
  registrar where supportertype = 'L';
  
  What I want to do is count the total number of records and then the 
  total number of records that have L as the supportertype 
 and then display them.
  
  So I should have something like There are 100 total records and 22 
  with Supporter Type L.
 
 Try this:
 
 SELECT COUNT(*), SUM(IF(supportertype = 'L', 1, 0)) FROM registrar;
 
 Baron
 
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Count syntax

2007-09-28 Thread Beauford
Hi,

I have the following line of code and I keep getting wrong results from it.
Can someone let me know what I'm doing wrong here. I just can't quite figure
out the syntax that I need.

select count(*) as numrows, count(supportertype) as leadcar from registrar
where supportertype = 'L';

What I want to do is count the total number of records and then the total
number of records that have L as the supportertype and then display them.

So I should have something like There are 100 total records and 22 with
Supporter Type L.

Thanks




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RE: Count syntax

2007-09-28 Thread Beauford
Thanks to all. 

 -Original Message-
 From: Michael Dykman [mailto:[EMAIL PROTECTED] 
 Sent: September 28, 2007 1:36 PM
 To: Beauford
 Cc: mysql@lists.mysql.com
 Subject: Re: Count syntax
 
 1 means that 1 will be added to the sum if the condition 
 tests, otherwise 0 will be added to the sum.  the condition 
 in this case is (supportertype = 'L') and will be applied to 
 every row.
 
 On 9/28/07, Beauford [EMAIL PROTECTED] wrote:
  Thanks - it works, but what does the 1 and 0 do in this - 
  SUM(IF(supportertype = 'L', 1, 0))
 
   -Original Message-
   From: Baron Schwartz [mailto:[EMAIL PROTECTED]
   Sent: September 28, 2007 1:00 PM
   To: Beauford
   Cc: mysql@lists.mysql.com
   Subject: Re: Count syntax
  
   Beauford wrote:
Hi,
   
I have the following line of code and I keep getting wrong
   results from it.
Can someone let me know what I'm doing wrong here. I just
   can't quite
figure out the syntax that I need.
   
select count(*) as numrows, count(supportertype) as 
 leadcar from 
registrar where supportertype = 'L';
   
What I want to do is count the total number of records and then 
the total number of records that have L as the supportertype
   and then display them.
   
So I should have something like There are 100 total 
 records and 
22 with Supporter Type L.
  
   Try this:
  
   SELECT COUNT(*), SUM(IF(supportertype = 'L', 1, 0)) FROM 
 registrar;
  
   Baron
  
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  - [EMAIL PROTECTED]
 
  - All models are wrong.  Some models are useful.
 
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Scheduled events

2007-08-14 Thread Beauford
Hi,

I have been trying for days to find a simple command in MySQL where I can
automatically delete records based on some criteria after a certain
timeframe.

I found this in the MySQL manual, but I guess it only works with version
5.1+. Unfortunately the version I need this to work on is 4.1, and can't be
upgraded.

CREATE EVENT e_hourly
ON SCHEDULE 
  EVERY 1 HOUR
COMMENT 'Clears out sessions table each hour.'
DO
  DELETE FROM something;

Is there something similar in version 4.1 that will do the same thing.

Thanks



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RE: Scheduled events

2007-08-14 Thread Beauford
  I have been trying for days to find a simple command in 
 MySQL where I 
  can automatically delete records based on some criteria after a 
  certain timeframe.
  
  I found this in the MySQL manual, but I guess it only works with 
  version 5.1+. Unfortunately the version I need this to work 
 on is 4.1, 
  and can't be upgraded.
  
  CREATE EVENT e_hourly
  ON SCHEDULE 
EVERY 1 HOUR
  COMMENT 'Clears out sessions table each hour.'
  DO
DELETE FROM something;
  
  Is there something similar in version 4.1 that will do the 
 same thing.
 
 No.  But there are cron jobs :-)  And if you're deleting many 
 rows and you don't want to interfere with other running 
 queries (sounds like this is an OLTP system), try MySQL 
 Archiver with the --purge argument 
 (http://mysqltoolkit.sourceforge.net/).  It's much more 
 efficient at large data volumes than a single DELETE statement.
 
 Baron
 

Really. I thought it would have some kind of scripting capability. I did
check out the link, but really don't need anything that extensive.

Is there a way to run the following command via cron.

delete from contacts where TO_DAYS(CURDATE()) - TO_DAYS(today) = 30 and
status != Y;

Thanks



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RE: Windows - logging into MySQL

2005-11-30 Thread Beauford
This is why lists like this get a bad name. First off, these instructions
are for 4.1. Secondly. What makes you think I haven't read the instructions
or searched extensively on Google etc. Either give a proper answer or just
shut your mouth. Never mind anyone responding, I'll find my own damn
answers. Thanks for the help.



-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] 
Sent: November 30, 2005 12:16 AM
To: Beauford
Cc: mysql@lists.mysql.com
Subject: Re: Windows - logging into MySQL



Beauford [EMAIL PROTECTED] wrote on 11/29/2005 11:31:24 PM:

 I just installed MySQL 3.23 (the only one I could get to work) and I 
 find that I can log into the server just by typing mysql at the DOS 
 prompt. Ho do make it so you have to use a user name and password to get
in.
 
 Thanks
 
 Running on Win2k Advance Server.
 
 
 

You could try following the installation instructions in the manual 
http://dev.mysql.com/doc/refman/4.1/en/windows-installation.html 

Shawn Green
Database Administrator
Unimin Corporation - Spruce Pine




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RE: Windows - logging into MySQL

2005-11-30 Thread Beauford
I have already changed the PW for root and I can log in no problem with
mysql -p root, but I can also get in with just mysql. Someone mentioned
about anonymous access, I'll have to check on that. 4.0 and above give me an
error that permission is denied for [EMAIL PROTECTED] I am not using an
external database so not sure why this is.

Thanks to all that posted legitimate answers. SGreen can biteme however. 

-Original Message-
From: Cal Evans [mailto:[EMAIL PROTECTED] 
Sent: November 30, 2005 12:08 AM
To: Beauford
Cc: mysql@lists.mysql.com
Subject: Re: Windows - logging into MySQL

Hi,

If you've just installed it then the root user has no password.
Change the root user to have a password then you will have to specify the -p
parameter on the command line before you can log in. (Or you can put it in
your local my.cnf file)

=C=
www.calevans.com


Beauford wrote:
 I just installed MySQL 3.23 (the only one I could get to work) and I 
 find that I can log into the server just by typing mysql at the DOS 
 prompt. Ho do make it so you have to use a user name and password to get
in.
 
 Thanks
 
 Running on Win2k Advance Server.
 
 
 

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RE: Windows - logging into MySQL

2005-11-30 Thread Beauford
There are always idots like Shawn Green that give BS answers just for the
sake of responding, and for me it just reflects badly on the group as a
whole. There are many others that help, but when I open my email and this is
the first response I see, it just gets under my skin. 

FYI. I have posted some information in another response as to the problems
with later versions etc.

-Original Message-
From: Logan, David (SST - Adelaide) [mailto:[EMAIL PROTECTED] 
Sent: November 30, 2005 5:57 PM
To: Beauford; [EMAIL PROTECTED]
Cc: mysql@lists.mysql.com
Subject: RE: Windows - logging into MySQL

Sorry Beauford,

The url that Shawn gave you is valid for the version you quoted. There
hasn't been much change and they work just as well. This is also the
earliest documentation that now exists on the website.

What is the bad name this list has gained? I've found it helpful on numerous
occasions. I think a reconsideration of your opinions would result in a
change of heart.

The documentation states the defaults that the mysql client program would
accept, it also details how to set users and passwords exactly as you
requested. These all work on 3.23.58 (I have this version installed at a
customer site as we write). Perhaps if you detail the problems you had
installing the later versions, somebody may be able to help.

Regards

David Logan
Database Administrator
HP Managed Services
148 Frome Street,
Adelaide 5000
Australia

+61 8 8408 4273 - Work
+61 417 268 665 - Mobile
+61 8 8408 4259 - Fax


-Original Message-
From: Beauford [mailto:[EMAIL PROTECTED]
Sent: Thursday, 1 December 2005 9:21 AM
To: [EMAIL PROTECTED]
Cc: mysql@lists.mysql.com
Subject: RE: Windows - logging into MySQL

This is why lists like this get a bad name. First off, these instructions
are for 4.1. Secondly. What makes you think I haven't read the instructions
or searched extensively on Google etc. Either give a proper answer or just
shut your mouth. Never mind anyone responding, I'll find my own damn
answers. Thanks for the help.



-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: November 30, 2005 12:16 AM
To: Beauford
Cc: mysql@lists.mysql.com
Subject: Re: Windows - logging into MySQL



Beauford [EMAIL PROTECTED] wrote on 11/29/2005 11:31:24 PM:

 I just installed MySQL 3.23 (the only one I could get to work) and I 
 find that I can log into the server just by typing mysql at the DOS 
 prompt. Ho do make it so you have to use a user name and password to
get
in.
 
 Thanks
 
 Running on Win2k Advance Server.
 
 
 

You could try following the installation instructions in the manual 
http://dev.mysql.com/doc/refman/4.1/en/windows-installation.html 

Shawn Green
Database Administrator
Unimin Corporation - Spruce Pine




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RE: Windows - logging into MySQL

2005-11-30 Thread Beauford
Sorry, but I don't take kindly to idiots. I don't care if he's gods gift to
MySQL. If there isn't enough info for him to give an informed answer then he
should either ask for more or shut up. I'm just simply trying to resolve a
problem, and flip answers aren't needed or appropriate.

I don't send emails to the list at the first sign of problem, this is
probably the first in over a year and a half. When I do however, it is
because the manual or other attempts have not provided me the answers I
need. 

In any event I will move on and find my answers elsewhere.

-Original Message-
From: J.R. Bullington [mailto:[EMAIL PROTECTED] 
Sent: November 30, 2005 6:37 PM
To: mysql@lists.mysql.com
Subject: RE: Windows - logging into MySQL

You need to be careful what you write. I have found that Shawn is one of the
best guys on this list. His answers are 99% right on target (1% because not
enough information was given) and can help on every version of db out there.

Don't start a flame war because your feelings were hurt. Grow up.

J.R.

-Original Message-
From: Beauford [mailto:[EMAIL PROTECTED]
Sent: Wednesday, November 30, 2005 6:05 PM
To: 'Logan, David (SST - Adelaide)'
Cc: mysql@lists.mysql.com
Subject: RE: Windows - logging into MySQL

There are always idots like Shawn Green that give BS answers just for the
sake of responding, and for me it just reflects badly on the group as a
whole. There are many others that help, but when I open my email and this is
the first response I see, it just gets under my skin. 

FYI. I have posted some information in another response as to the problems
with later versions etc.

-Original Message-
From: Logan, David (SST - Adelaide) [mailto:[EMAIL PROTECTED]
Sent: November 30, 2005 5:57 PM
To: Beauford; [EMAIL PROTECTED]
Cc: mysql@lists.mysql.com
Subject: RE: Windows - logging into MySQL

Sorry Beauford,

The url that Shawn gave you is valid for the version you quoted. There
hasn't been much change and they work just as well. This is also the
earliest documentation that now exists on the website.

What is the bad name this list has gained? I've found it helpful on numerous
occasions. I think a reconsideration of your opinions would result in a
change of heart.

The documentation states the defaults that the mysql client program would
accept, it also details how to set users and passwords exactly as you
requested. These all work on 3.23.58 (I have this version installed at a
customer site as we write). Perhaps if you detail the problems you had
installing the later versions, somebody may be able to help.

Regards

David Logan
Database Administrator
HP Managed Services
148 Frome Street,
Adelaide 5000
Australia

+61 8 8408 4273 - Work
+61 417 268 665 - Mobile
+61 8 8408 4259 - Fax


-Original Message-
From: Beauford [mailto:[EMAIL PROTECTED]
Sent: Thursday, 1 December 2005 9:21 AM
To: [EMAIL PROTECTED]
Cc: mysql@lists.mysql.com
Subject: RE: Windows - logging into MySQL

This is why lists like this get a bad name. First off, these instructions
are for 4.1. Secondly. What makes you think I haven't read the instructions
or searched extensively on Google etc. Either give a proper answer or just
shut your mouth. Never mind anyone responding, I'll find my own damn
answers. Thanks for the help.



-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: November 30, 2005 12:16 AM
To: Beauford
Cc: mysql@lists.mysql.com
Subject: Re: Windows - logging into MySQL



Beauford [EMAIL PROTECTED] wrote on 11/29/2005 11:31:24 PM:

 I just installed MySQL 3.23 (the only one I could get to work) and I 
 find that I can log into the server just by typing mysql at the DOS 
 prompt. Ho do make it so you have to use a user name and password to
get
in.
 
 Thanks
 
 Running on Win2k Advance Server.
 
 
 

You could try following the installation instructions in the manual 
http://dev.mysql.com/doc/refman/4.1/en/windows-installation.html 

Shawn Green
Database Administrator
Unimin Corporation - Spruce Pine




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Windows - logging into MySQL

2005-11-29 Thread Beauford
I just installed MySQL 3.23 (the only one I could get to work) and I find
that I can log into the server just by typing mysql at the DOS prompt. Ho do
make it so you have to use a user name and password to get in.

Thanks

Running on Win2k Advance Server.



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RE: Problem with tables

2004-09-16 Thread Beauford
Duh.That's what I get for doing this stuff late at night. Funning how
some things just don't click in until someone kicks you in the butt

Thanks 

-Original Message-
From: Aman Raheja [mailto:[EMAIL PROTECTED] 
Sent: September 16, 2004 8:35 AM
To: [EMAIL PROTECTED]
Subject: Re: Problem with tables;

You need to do
desc spamdata;
NOT desc id.

Aman Raheja


Beauford wrote:

Hi,

I'm trying to create a table using the commands below, and everything 
seems to work and the table is there - I just can's access any of the 
fields. This happens on both my Windows and Linux versions of MySQL. 
Any help is appreciated. Note, I have 8 other databases that I created 
with no problem about 2 months ago (on both platforms) - this is the 
first time I have tried since then. Any help is appreciated

Thanks

I am in the proper database when I run the commands below...

CREATE TABLE spamdata (
 id  bigint(20)   NOT NULL,
 dates   varchar(60)  NOT NULL,
 probe   varchar(100) NOT NULL,
 ip  varchar(15)  NOT NULL,
 fqdnvarchar(100) NOT NULL,
 reason  varchar(100) NOT NULL,
 portvarchar(2)   NOT NULL,
 codevarchar(10)  NOT NULL,
PRIMARY KEY (idm));

Query OK, 0 rows affected (0.08 sec) *** This tells me nothing happened...

When I run 'desc id' (or any of the others) I get the following:

mysql desc id;
ERROR 1146: Table 'spamtool.id' doesn't exist

This shows the table is there though.

mysql show tables;
++
| Tables_in_spamtool |
++
| spamdata   |
++
1 rows in set (0.00 sec)


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Problem with tables;

2004-09-15 Thread Beauford
Hi,

I'm trying to create a table using the commands below, and everything seems
to work and the table is there - I just can's access any of the fields. This
happens on both my Windows and Linux versions of MySQL. Any help is
appreciated. Note, I have 8 other databases that I created with no problem
about 2 months ago (on both platforms) - this is the first time I have tried
since then. Any help is appreciated

Thanks

I am in the proper database when I run the commands below...

CREATE TABLE spamdata (
 id  bigint(20)   NOT NULL,
 dates   varchar(60)  NOT NULL,
 probe   varchar(100) NOT NULL,
 ip  varchar(15)  NOT NULL,
 fqdnvarchar(100) NOT NULL,
 reason  varchar(100) NOT NULL,
 portvarchar(2)   NOT NULL,
 codevarchar(10)  NOT NULL,
PRIMARY KEY (idm));

Query OK, 0 rows affected (0.08 sec) *** This tells me nothing happened...

When I run 'desc id' (or any of the others) I get the following:

mysql desc id;
ERROR 1146: Table 'spamtool.id' doesn't exist

This shows the table is there though.

mysql show tables;
++
| Tables_in_spamtool |
++
| spamdata   |
++
1 rows in set (0.00 sec)




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Problem with tables;

2004-09-15 Thread Beauford
Just some updated info on this. It appears everything is there and working.
I can update the table, do queries on the table, etc. - the only thing I
can't do is a 'desc whatever' I still get the 1146 error.

Thanks 

-Original Message-
From: Beauford [mailto:[EMAIL PROTECTED] 
Sent: September 16, 2004 12:19 AM
To: [EMAIL PROTECTED]
Subject: Problem with tables;

Hi,

I'm trying to create a table using the commands below, and everything seems
to work and the table is there - I just can's access any of the fields. This
happens on both my Windows and Linux versions of MySQL. Any help is
appreciated. Note, I have 8 other databases that I created with no problem
about 2 months ago (on both platforms) - this is the first time I have tried
since then. Any help is appreciated

Thanks

I am in the proper database when I run the commands below...

CREATE TABLE spamdata (
 id  bigint(20)   NOT NULL,
 dates   varchar(60)  NOT NULL,
 probe   varchar(100) NOT NULL,
 ip  varchar(15)  NOT NULL,
 fqdnvarchar(100) NOT NULL,
 reason  varchar(100) NOT NULL,
 portvarchar(2)   NOT NULL,
 codevarchar(10)  NOT NULL,
PRIMARY KEY (idm));

Query OK, 0 rows affected (0.08 sec) *** This tells me nothing happened...

When I run 'desc id' (or any of the others) I get the following:

mysql desc id;
ERROR 1146: Table 'spamtool.id' doesn't exist

This shows the table is there though.

mysql show tables;
++
| Tables_in_spamtool |
++
| spamdata   |
++
1 rows in set (0.00 sec)




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Searh question

2003-10-06 Thread Beauford
If I have a database with various dates in it, how can I search for all
instances of the same date. Note that I do not know what the dates are
(or how many there will be) so the search can't be hardcoded. These
dates will be added based on user input from a website. 

Is there some kind of comparison function where I can search on field
that have the same values..?

TIA

PS I want to end up with a table that follows this format (this is
just an example)

01/01/03

Name Department Title
Name Department Title
Name Department Title
Name Department Title
Name Department Title

05/01/03

Name Department Title
Name Department Title
Name Department Title
Name Department Title
Name Department Title


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Help. Not sure my other email got through!

2003-02-09 Thread Beauford
Not sure my other email got through, so here it is again, with an updated
description of the problem.

Hi,

 I have a database with 3 names in it. In each of these names is 5
categories  that have numbers in them. I want to be able to do a select and
total these numbers for each name, and then sort them in descending order. I
have tried for over 6 hours tonight and no matter what I do, there is one
minor glitch. I am not supplying any code as I would like to know from you
people first what the best way to go about this would be - as what I have
done is
obviously not working.

 TIA

 i.e

This is what is in the database.

Table 1

ID1 NAME

1  Bill
2  John
3  Fred

Table 2

ID2  Col1 Col2 Col3 Col4

1   1456
2   3286
3   4922

Database 3 is a reference file that ties them together - so I would have to
use a join in the select statement.

ID1  ID2
11
22
33

This is what I want to be able to show:

John   21
Fred   20
Bill 16





























sql, query, queries, smallint



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Linux Mysql vs Windows Mysql

2003-01-14 Thread Beauford
Hi,

I just installed MySQL and PHP on Windows XP and have a question. I created
a test database on Windows (exactly the same as the test one I use in Linux)
and copied a test PHP script which accesses Mysql and displays some
information. On Linux it works perfectly, but on Windows I get the following
error. Are there different commands or syntax on the two different
platforms?

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result
resource in e:\directory\index.php on line 21.

This is the exact code below:

?
PRINT HTML;
PRINT HEAD;
PRINT TITLEADDRESSES/title;

PRINT /HEAD;

PRINT BODY BGCOLOR=white TEXT=#00 LINK=#FF VLINK=#66
aLINK=#FF;

$out[]=0;   // I also got an error about this - without it here I get an
undefined variable error below

mysql_connect(localhost:3308, user, password) or
die(could not connect);
mysql_select_db(address);

$query = SELECT * FROM data;
$result = mysql_query ($query);

 while ($line = mysql_fetch_row($result)) {

$out[]=$i;

}

?

form action=? echo $PHP_SELF; ? method=post

PTABLE WIDTH=400 BORDER=0


TRTD ALIGN=leftName :/TDTD ALIGN=leftINPUT TYPE=text
NAME=lastname SIZE=20 VALUE=? echo
$out[0][1]?/TD/TR
TRTD ALIGN=leftName :/TDTD ALIGN=leftINPUT TYPE=text
NAME=lastname SIZE=20 VALUE=? echo
$out[0][2]?/TD/TR
TRTD ALIGN=leftCity:/TDTD ALIGN=leftINPUT TYPE=text
NAME=city SIZE=20 VALUE=? echo
$out[0][3] ?/TD/TR

/TABLE

/FORM




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Duplicating a table question

2002-12-14 Thread Beauford.2003
These spam filters are really annoying. It's hard to get an answer to my
question when my message gets bounced 'cause it doesn't have MySQL or query
in the message.

In any event,

 Is there a way to duplicate a complete table. I am making some
modifications and don't want to use the working table for testing.

 TIA



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Automatically totalling columns.

2002-12-14 Thread Beauford
Hi,

I have two tables, one with peoples names in it and one with data associated
to these people. What I need to do is to be able to automatically total
certain items in the data table that correspond with a name in the first
table and then update the name table..

i.e. This is a crude example, but you get the picture.  When I add an item
to the data table I want to have it so it automatically gets added to the
correct person in the name table. I also have a third table which references
these two tables. Can this be done from within MySQL.

Names  Data

Col. 1  Col 2 Col. 1Col. 2

Fred$2.24Milk$1.00
Wilma  $4.25Bread $1.24
Barney $1.00Eggs   $2.19
Betty$3.19Butter $3.25




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Re: Load Data Infile command

2002-12-12 Thread Beauford.2003
Try this when your logging into mysql, not when starting mysql. This works
for me on Linux.

mysql --local-infile=1 -u username-p

- Original Message -
From: Jon Bertsch [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, December 12, 2002 12:02 PM
Subject: Load Data Infile command


 HI all,

 I'm new to the list so I apologize if this subject has been answered
 already, I couldn't find a mailing list archive.

 We just upgraded (?) mysql to 3.23.49. We are running it on Linux Redhat
7.3

 The load data local command is turned off by default. I have attempted to
 follow the instructions in the documentation (Ch. 4.2.4 and others) about
 this but it has been a dismal failure.

 I tried the following on restarting mysql:

 /etc/init.d/mysqld start --local-infile[=1]

 as in the manual. No luck.

 I also made some variations on this:

 --local-infile=1
 --local-infile
 --local-infile(=1)
 --local-infile=[1]
 --local-infile=(1)

 None of these worked.

 I added the following two lines to the my.cnf file:

 [client]
 enable-local-infile

 and restarted. No luck.

 Has anyone been able to get this to work? I would really appreciate any
 help on this since I use the command to load data and it is part of
several
 utility applications I use.

 Thanks in advance

 Jon Bertsch


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Displaying output from MySQL

2002-12-10 Thread Beauford.2003
Hi,

Not sure if this is a PHP of a MySQL question, so I am sending it to both
groups. Basically I have a list of numbers with two decimal places in the
MySQL database, but I only want to display some of them with the decimal
points.

i.e.

70 (not 70.00)
87
51.5
46.75
12
29
5.5
-1
45

I know it's probably a weird request, but any thoughts on how one would do
this either through PHP or MySQL.

TIA



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Re: Select * From table where name Like 'help'; Help

2002-12-09 Thread Beauford.2003
I am using PHP on my website, but this is certainly a MySQL question.

 That shouldn't matter, because the real value $var is inserted before
 the query is send to the MySQL server, where REGEXP of the query is
 evaluated. If there is a problem, you need to be much more
 specific. Quote the error message, show the relevant part of your code
 and so on.

The real value of $var is not inserted before it is sent to the server.
REGEXP thinks that 'var' is part of the search pattern because $ is a
reserved operator for REGEXP. So what gets sent to the server is totally
different than what I want to be sent. There are no errors because the
syntax is correct, its just not correct for the search I want it to perform.
So the questions still remains, how do I get REGEXP to treat $var as a
variable and not part of its own syntax.

I have been looking at http://www.mysql.com/doc/en/Regexp.html, but it does
not cover this.

- Original Message -
From: Benjamin Pflugmann [EMAIL PROTECTED]
To: Beauford.2003 [EMAIL PROTECTED]
Cc: Robert Citek [EMAIL PROTECTED]; MySQL List
[EMAIL PROTECTED]
Sent: Monday, December 09, 2002 3:30 AM
Subject: Re: Select * From table where name Like 'help'; Help


 Hello.

 On Mon 2002-12-09 at 01:00:33 -0500, [EMAIL PROTECTED] wrote:
  I am doing this search from a website,

 What does that mean? Which scripting language do you use? PHP? This is
 more a question for your website environment than about MySQL.

  and from what I can tell there is no way to use a variable
  (i.e. $var) with REGEXP as the $ sign has a special meaning to
  REGEXP.

 That shouldn't matter, because the real value $var is inserted before
 the query is send to the MySQL server, where REGEXP of the query is
 evaluated. If there is a problem, you need to be much more
 specific. Quote the error message, show the relevant part of your code
 and so on.

 HTH,

 Benjamin.

 --
 [EMAIL PROTECTED]

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Re: Select * From table where name Like 'help'; Help

2002-12-09 Thread Beauford.2003
My mistake, I assumed the space was just for clarity and not really
meaningful, but since it is between the ' ' it makes sense that is. I tried
a few variations on this and think I have found what I need - where title
like binary '%$var%.  Notice the key word 'binary' which makes the world of
difference.

From the tests I have run this will find the occurrence of any word that
begins with a capital and then matches any remaining characters. So if I
search for One it will fine One Day, Day One, This One Day - however, if I
search for On it will also find On The Water and Your On Third.

The only thing I could do is to add a space after $var, but them partial
searches won't be found. i.e. searching for beaufor will not find beauford.

Until I can find a better way, one of these will suffice.

Thanks for your correction as I did miss the space thing.

Beauford


- Original Message -
From: Jocelyn Fournier [EMAIL PROTECTED]
To: Beauford.2003 [EMAIL PROTECTED]
Sent: Monday, December 09, 2002 9:31 AM
Subject: Re: Select * From table where name Like 'help'; Help


 No it doesn't, since I introduce a space between % and one. So stone
will
 not be returned :)

 Regards,
   Jocelyn
 - Original Message -
 From: Beauford.2003 [EMAIL PROTECTED]
 To: Jocelyn Fournier [EMAIL PROTECTED]
 Sent: Monday, December 09, 2002 2:26 PM
 Subject: Re: Select * From table where name Like 'help'; Help


  That is what I am using, however, it is not as detailed as I need it to
 be.
  Like I said in my original question, I don't want to get 'stone' when I
  search for 'one', and using LIKE does exactly this.
 
  Thanks, Beauford
 
 
  - Original Message -
  From: Jocelyn Fournier [EMAIL PROTECTED]
  To: Beauford.2003 [EMAIL PROTECTED]; MySQL List
  [EMAIL PROTECTED]
  Sent: Sunday, December 08, 2002 7:57 PM
  Subject: Re: Select * From table where name Like 'help'; Help
 
 
   Hi,
  
   A dirty solution would be to search :
  
   field LIKE '% one %' or field LIKE 'one %' or field LIKE '% one'  or
   field='one';
  
   Regards,
 Jocelyn
  
   - Original Message -
   From: Beauford.2003 [EMAIL PROTECTED]
   To: MySQL List [EMAIL PROTECTED]
   Sent: Tuesday, December 10, 2002 12:50 AM
   Subject: Select * From table where name Like 'help'; Help
  
  
Hi,
   
Hey, did this list change their spam filters? Now every time I send
an
   email
it gets bounced unless the word MySQL  or Query is in the email.
 Anyway,
Please read below for a recent problem I am having.
   
 I want to be able to search on my database and be able to get the
following, but running into a few snags.
   
 If I search for the word - One - I want to see:
   
 Another One
One Day
One on One
Your The One
   
 But not:
   
 Fashioned
Stone
Everyone
   
 Also, and maybe part of the solution - is there a way for the
search
 to
   be
case sensitive.
   
 TIA
   
 Beauford
   
   
   
   
   
   
   
  
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Re: Select * From table where name Like 'help'; Help

2002-12-09 Thread Beauford.2003
Andy,

I mentioned in my previous email that I am using PHP, and I have also tried
putting quotes around $var (many different ways) with no better results.
REGEXP just gives a syntax error when I do this.

Through the suggestion of another list user I have found a way that suits my
needs using the where name like binary '$var'. It's jury rigged, but works
the way I need it to work for the most part.

Thanks for your suggestions.

- Original Message -
From: Andy Bakun [EMAIL PROTECTED]
To: Beauford.2003 [EMAIL PROTECTED]
Cc: MySQL List [EMAIL PROTECTED]
Sent: Monday, December 09, 2002 3:08 PM
Subject: Re: Select * From table where name Like 'help'; Help


 You still have not mentioned what language you are using to interface
 with MySQL.  If you are using PHP or Perl, then things like single and
 double quotes will make a difference here, and looking in the MySQL
 manual will not help you.  Check the string that contains the query, if
 it looks like this:

$q = 'select  from table where col regexp $var';

 change it to

$q = select  from table where col regexp '$var';

 The enclosing single quotes are keeping the value of $var from being
 interpolated into the string in the above example.  See the manual for
 your programming language for more information on variable
 interpolation.

 On Mon, 2002-12-09 at 08:22, Beauford.2003 wrote:
  I am using PHP on my website, but this is certainly a MySQL question.
 
   That shouldn't matter, because the real value $var is inserted before
   the query is send to the MySQL server, where REGEXP of the query is
   evaluated. If there is a problem, you need to be much more
   specific. Quote the error message, show the relevant part of your code
   and so on.
 
  The real value of $var is not inserted before it is sent to the server.
  REGEXP thinks that 'var' is part of the search pattern because $ is a
  reserved operator for REGEXP. So what gets sent to the server is totally
  different than what I want to be sent. There are no errors because the
  syntax is correct, its just not correct for the search I want it to
perform.
  So the questions still remains, how do I get REGEXP to treat $var as a
  variable and not part of its own syntax.
 
  I have been looking at http://www.mysql.com/doc/en/Regexp.html, but it
does
  not cover this.
 
  - Original Message -
  From: Benjamin Pflugmann [EMAIL PROTECTED]
  To: Beauford.2003 [EMAIL PROTECTED]
  Cc: Robert Citek [EMAIL PROTECTED]; MySQL List
  [EMAIL PROTECTED]
  Sent: Monday, December 09, 2002 3:30 AM
  Subject: Re: Select * From table where name Like 'help'; Help
 
 
   Hello.
  
   On Mon 2002-12-09 at 01:00:33 -0500, [EMAIL PROTECTED] wrote:
I am doing this search from a website,
  
   What does that mean? Which scripting language do you use? PHP? This is
   more a question for your website environment than about MySQL.
  
and from what I can tell there is no way to use a variable
(i.e. $var) with REGEXP as the $ sign has a special meaning to
REGEXP.
  
   That shouldn't matter, because the real value $var is inserted before
   the query is send to the MySQL server, where REGEXP of the query is
   evaluated. If there is a problem, you need to be much more
   specific. Quote the error message, show the relevant part of your code
   and so on.
  
   HTH,
  
   Benjamin.
  
   --
   [EMAIL PROTECTED]
  
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Re: Select * From table where name Like 'help'; Help

2002-12-09 Thread Beauford.2003

 This is most definitely a quoting issue with your PHP string.  If it is
 quoted correctly, the MySQL server will never see the string $var
 appearing after the keyword REGEXP, so if $ has meaning to a regular
 expression wouldn't matter (unless the value of $var contained a $).
 I've used REGEXP with (really complex) dynamicly built regular
 expressions stored in PHP variables a number of times and have never had
 a problem as you've described above.

MySQL is not seeing $var, but it is seeing var from what I can understand of
Regexp (the manual on this is not very good), $ means to match the end of a
string, so since I have $var, I am thinking that Regexp thinks that the end
of the string I want to search for ends with var. In any event, the actual
value of  $var never gets sent to MySQL, so the search will always fail.

Maybe I am misunderstanding the meaning of this, but since I really can't
make any sense out of the manual, I am guessing at what I think they mean.

Let me give you and example:

$var=One ($var is passed the string from a form on a webpage)

There are 3 entries in the database with the word One

One Day
Day One
That's the One

There are also other words that contain the word one, like stone and anyone.

I want to be able to run a search that will just show the first 3 by typing
in the word One ('one' would result in the string not found').

Beauford



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Fw: Thanks

2002-12-08 Thread Beauford.2003
Thanks to all who recently answered some of my MySQL questions, although I
am still having some problems with the concept of joins, but what I have
figured out certainly does make it more efficient and versatile. I have
managed to eliminate two tables (actually combine them into one) which made
my searches easier.

Thanks again, Beauford




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Re: LOAD DATA command

2002-12-08 Thread Beauford.2003
If your using a Linux machine try using this when logging on to mysql:

mysql --local-infile=1 -u username -p

Not sure about Windows.

There is apparently a security issue with this so it is not enable by
default. Can't remember where I read this, but the above works for me.

Beauford

- Original Message -
From: Insanely Great [EMAIL PROTECTED]
To: Padma kuruganti [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Saturday, December 07, 2002 12:32 PM
Subject: Re: LOAD DATA command


 Greetings...

 Are you using Windows to connect to the MySQL. Then you can try SQLyog at
 http://www.webyog.com/sqyog/download.html It will help you a lot in these
 works and you dont have to write queries also.

 Rgds
 Insane

 - Original Message -
 From: Padma kuruganti [EMAIL PROTECTED]
 To: [EMAIL PROTECTED]
 Sent: Saturday, December 07, 2002 12:46 AM
 Subject: LOAD DATA command


  Hi I just downloaded version 3.23.53 for evaluation.
  when I try to use LOAD DATA LOCAL INFILE 'some.txt
  into TABLE sometable I get a message saying 'the used
  command is not allowed in this version of mySQL. IS
  there any update on this? I could not find any info on
  your web site.
  Thanks
  KP
 
  __
  Do you Yahoo!?
  Yahoo! Mail Plus - Powerful. Affordable. Sign up now.
  http://mailplus.yahoo.com
 
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Select * From table where name Like 'help'; Help

2002-12-08 Thread Beauford.2003
Hi,

Hey, did this list change their spam filters? Now every time I send an email
it gets bounced unless the word MySQL  or Query is in the email. Anyway,
Please read below for a recent problem I am having.

 I want to be able to search on my database and be able to get the
following, but running into a few snags.

 If I search for the word - One - I want to see:

 Another One
One Day
One on One
Your The One

 But not:

 Fashioned
Stone
Everyone

 Also, and maybe part of the solution - is there a way for the search to be
case sensitive.

 TIA

 Beauford







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Re: Select * From table where name Like 'help'; Help

2002-12-08 Thread Beauford.2003
I am doing this search from a website, and from what I can tell there is no
way to use a variable (i.e. $var) with REGEXP as the $ sign has a special
meaning to REGEXP. Any other suggestions, or would you know a way around
this.

TIA

- Original Message -
From: Robert Citek [EMAIL PROTECTED]
To: Beauford.2003 [EMAIL PROTECTED]
Cc: MySQL List [EMAIL PROTECTED]
Sent: Sunday, December 08, 2002 8:16 PM
Subject: Re: Select * From table where name Like 'help'; Help



 Try the REGEXP modifier.  For example:

 mysql select Another One regexp [[::]]One[[::]] ;
 mysql select Fashioned regexp [[::]]One[[::]] ;

 The format differs from that used in sed, grep, awk, perl, python, etc.


http://www.mysql.com/documentation/mysql/bychapter/manual_Regexp.html#Regexp

 Regards,
 - Robert

 -

 At 07:50 PM 12/9/2002 -0500, Beauford.2003 wrote:
  If I search for the word - One - I want to see:
 Another One
  But not:
 Fashioned


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Join help.

2002-12-04 Thread Beauford.2003
Hi,

Is there a good tutorial somewhere on the join command. No matter what I do
it just doesn't work. Obviously I'm doing something wrong, but the MySQL
manual just doesn't help at all.

TIA



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Re: Table setup question

2002-12-04 Thread Beauford.2003
Adolfo,

Thanks for the info, but can you elaborate on it, 'cause basically I am just
not getting the concept. No way no how can I get these joins to work.

mysql describe songs;  (this contains an ID field and the title of the song
and the artist.)
++-+--+-+-+---+
| Field  | Type| Null | Key | Default | Extra |
++-+--+-+-+---+
| ids| int(11) |  | PRI | 0   |   |
| title  | varchar(55) | YES  | | NULL|   |
| artist | varchar(30) | YES  | | NULL|   |
++-+--+-+-+---+
3 rows in set (0.00 sec)

mysql describe album;  (this contains an ID field and the name of the album
the above songs came from.)
+---+-+--+-+-+---+
| Field | Type| Null | Key | Default | Extra |
+---+-+--+-+-+---+
| ida   | int(11) |  | PRI | 0   |   |
| name  | varchar(35) | YES  | | NULL|   |
+---+-+--+-+-+---+
2 rows in set (0.00 sec)

mysql describe reference; (this contains the ID's that corrspond to the
above two tables.)
+---+-+--+-+-+---+
| Field | Type| Null | Key | Default | Extra |
+---+-+--+-+-+---+
| ida   | int(11) |  | PRI | 0   |   |
| ids   | int(11) |  | PRI | 0   |   |
+---+-+--+-+-+---+
2 rows in set (0.00 sec)

This table is in this format: The left column corresponds to the ID of the
ablum and the right field correspond to the ID of the song. From the example
below, you can see that song #1 appears on Albums 2, 3, 6, and 16.

SongAlbum
12
13
16
115
27
221
223

So with all this information, how would I perform a select that would show
the all the albums a particular song were on. (from the example above what
would I need to do to show the album name and song title for song ID # 1
above).

Sorry if this is long, I just wanted to make sure there is enough info.

TIA

Beauford

- Original Message -
From: Adolfo Bello [EMAIL PROTECTED]
To: 'Beauford.2003' [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Monday, December 02, 2002 7:33 AM
Subject: RE: Table setup question


 Try something like
 CREATE TABLE album( ida int primary key , title varchar(n) not null ) ;
 CREATE TABLE songs( ids intprimary key, song varchar(m) not null ) ;
 CREATE TABLE albumsongs(
 ida int not null,
 ids int not null,
 primary key(ida,ids),
 foreign key(ida) references album(ida),
 foreign key(ids) references songs(ids));

 Querying for albums which have a songs is something like:
 SELECT title,song FROM album INNER JOIN albumsongs ON ...
 INNER JOIN songs ON ... WHERE song='your song name';


  -Original Message-
  From: Beauford.2003 [mailto:[EMAIL PROTECTED]]
  Sent: Sunday, December 01, 2002 9:59 PM
  To: [EMAIL PROTECTED]
  Subject: Table setup question
 
 
  Hi,
 
  I have a database of albums by a group and I want to be able
  to search on this table to find out what songs are duplicated
  on what albums (there are
  36+ albums). The problem is how do I set up the database.
 
  Example:
 
  Album1 has 3 songs.  1.song, 2.song, 3.song, 4.song
  Album2 has 4 songs.  4.song, 5.song, 6.song, 3.song
  Album3 has 4 songs.  7.song, 8.song, 1.song, 3.song
 
  So 3.song appears on all 3 albums.
 
  Currently I have it set up with two tables as shown below,
  but I am thinking there has to be a better way to do this
  than to duplicate the name of the song three, four, or five
  times in the table.
 
  Table AlbumName
 
  AlbumID
  ==
  Album1   1
  Album2   2
  Album3   3
 
  Table SongTitle
 
  SongID
  === ==
  3.song  1
  3.song  2
  3.song  3
  7.song  3
  etc.
  etc.
 
  So basically my search is  - SELECT Album, Song FROM
  AlbumName, SongTitle WHERE AlbumName.ID=SongTitle.ID;
 
  Given the setup above, is there a way that I can put in the
  SongTitle.ID field that song appears on more than one album.
  Maybe something like:
 
  SongID
  === ==
  3.song  1, 2, 3
 
  But then what would my search be.
 
  Sorry for the length of this, but I am learning MySQL and
  trying to get a handle on all of it. My way works, but I'm
  sure there has to be a better way.
 
  Any thoughts are appreciated.
 
  TIA, Beauford
 
 
 
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Re: Table setup question

2002-12-04 Thread Beauford.2003
Thanks for the help, but I just can't grasp the concept (I'm not even sure I
have my table layouts the way they should be to have this work). I'll have
to look into this further, but for the time being the way I have it is
working.

Beauford

- Original Message -
From: Adolfo Bello [EMAIL PROTECTED]
To: 'Beauford.2003' [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Wednesday, December 04, 2002 4:28 PM
Subject: RE: Table setup question


 Let's assume you want the entire albums list which the song
 Day and Night by Billie Holiday is in.

 SELECT t1.title,t1.artist,t3.name FROM sings t1
 INNER JOIN reference t2 ON t1.ids=t2.ids
 INNER JOIN album t3 ON t2.ida=t3.ida
 WHERE t1.title='Day and Night' AND t1.artist='Billie Holiday'

 (or WHERE t1.ids=1 if you know the id of the song)

 Relationship is like:

 [songs] [album]
 \\//
 [reference]
 (this later one contains the list of songs of each album)

 This way you treat songs and album separately and link them
 through another table so you don't have any limit on the number
 of albums a song can belong to.

 If you know the id of a song all you have to do is

 Adolfo

  -Original Message-
  From: Beauford.2003 [mailto:[EMAIL PROTECTED]]
  Sent: Thursday, December 05, 2002 3:42 PM
  To: Adolfo Bello
  Cc: [EMAIL PROTECTED]
  Subject: Re: Table setup question
 
 
  Adolfo,
 
  Thanks for the info, but can you elaborate on it, 'cause
  basically I am just not getting the concept. No way no how
  can I get these joins to work.
 
  mysql describe songs;  (this contains an ID field and the
  title of the
  mysql song
  and the artist.)
  ++-+--+-+-+---+
  | Field  | Type| Null | Key | Default | Extra |
  ++-+--+-+-+---+
  | ids| int(11) |  | PRI | 0   |   |
  | title  | varchar(55) | YES  | | NULL|   |
  | artist | varchar(30) | YES  | | NULL|   |
  ++-+--+-+-+---+
  3 rows in set (0.00 sec)
 
  mysql describe album;  (this contains an ID field and the
  name of the
  mysql album
  the above songs came from.)
  +---+-+--+-+-+---+
  | Field | Type| Null | Key | Default | Extra |
  +---+-+--+-+-+---+
  | ida   | int(11) |  | PRI | 0   |   |
  | name  | varchar(35) | YES  | | NULL|   |
  +---+-+--+-+-+---+
  2 rows in set (0.00 sec)
 
  mysql describe reference; (this contains the ID's that
  corrspond to the
  above two tables.)
  +---+-+--+-+-+---+
  | Field | Type| Null | Key | Default | Extra |
  +---+-+--+-+-+---+
  | ida   | int(11) |  | PRI | 0   |   |
  | ids   | int(11) |  | PRI | 0   |   |
  +---+-+--+-+-+---+
  2 rows in set (0.00 sec)
 
  This table is in this format: The left column corresponds to
  the ID of the ablum and the right field correspond to the ID
  of the song. From the example below, you can see that song #1
  appears on Albums 2, 3, 6, and 16.
 
  SongAlbum
  12
  13
  16
  115
  27
  221
  223
 
  So with all this information, how would I perform a select
  that would show the all the albums a particular song were on.
  (from the example above what would I need to do to show the
  album name and song title for song ID # 1 above).
 
  Sorry if this is long, I just wanted to make sure there is
  enough info.
 
  TIA
 
  Beauford
 
  - Original Message -
  From: Adolfo Bello [EMAIL PROTECTED]
  To: 'Beauford.2003' [EMAIL PROTECTED];
  [EMAIL PROTECTED]
  Sent: Monday, December 02, 2002 7:33 AM
  Subject: RE: Table setup question
 
 
   Try something like
   CREATE TABLE album( ida int primary key , title varchar(n)
  not null )
   ; CREATE TABLE songs( ids intprimary key, song varchar(m)
  not null ) ;
   CREATE TABLE albumsongs( ida int not null,
   ids int not null,
   primary key(ida,ids),
   foreign key(ida) references album(ida),
   foreign key(ids) references songs(ids));
  
   Querying for albums which have a songs is something like: SELECT
   title,song FROM album INNER JOIN albumsongs ON ... INNER
  JOIN songs ON
   ... WHERE song='your song name';
  
  
-Original Message-
From: Beauford.2003 [mailto:[EMAIL PROTECTED]]
Sent: Sunday, December 01, 2002 9:59 PM
To: [EMAIL PROTECTED]
Subject: Table setup question
   
   
Hi,
   
I have a database of albums by a group and I want to be able to
search on this table to find out what songs are
  duplicated on what
albums (there are
36+ albums). The problem is how do I set up the database.
   
Example:
   
Album1 has 3 songs.  1.song, 2.song, 3.song, 4.song
Album2 has 4 songs

Select, mysql_fetch_array, PHP question

2002-12-03 Thread Beauford.2003
Hi,

I'm trying to do a search on database through a web interface using PHP and
having one small problem. Example: If I search for items that cost $5.00 and
there are three of them, the query works and displays the information, but
if there is only one item the query works but doesn't display the item. I
just get a blank page.

SELECT price, item FROM list WHERE price=$5.00;

 while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {

 echo $item;  }

If I do the same search from the MySQL command line the query will display
one or more items.

Ideas?

TIA



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Re: Table setup question

2002-12-03 Thread Beauford.2003
DL,

Yep, I see what your saying. In my application it may work fine, but in
larger applications where things may change, it would be easier to do it
with the 3 tables.

Thanks

- Original Message -
From: DL Neil [EMAIL PROTECTED]
To: Beauford.2003 [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Tuesday, December 03, 2002 7:56 AM
Subject: Re: Table setup question


 Beauford,

  OK, I get it now. I thought there would need to be duplication in the
  AlbumSonglist DB.

 =indeed, the process of 'normalisation' (as mentioned earlier) helps to
 ascertain when and where 'duplication' is allowed/beneficial/should not
 happen.

 Just one other question though. What is the
  difference/benefits of doing it this way, than the way I have it.
 Currently
  I use two tables - the songlist table includes 2 id fields (one that
  corresponds with the album title, and one that is used for the song
 title).
  The difference between what you have said and what I have is that I have
  duplicated the actual song titles instead of a pointer to the song title
  This way I have eliminated one table.

 =try wording a query to get from album to song title, and then another
query
 starting from song title and working to album. That's always a good test
of
 two-way relationship situations.

 =if you have a pointer repeated then (usually) it will take up less
storage
 space than a repeated song title. OTOH getting rid of a table and the join
 in each query will speed up processing time/throughput (by a v.small
 amount)...

 =if some data ever changes (admittedly I can't see that it would in your
 application - but you're the subject matter expert), eg someone remixes a
 song, so to be more precise you want to modify Leila to have Leila -
 original and Leila - remix as two different entries in the db; then how
 easy would it be to go through the song title data and make that change?
In
 the m-m model the change would be needed only once (the Song Title tbl -
all
 else unaffected). In your model, you would need to be sure that you
changed
 every case - and without knowing how many changes is 'enough'! (now apply
 this to customer records and people changing addresses and you'll see
 why/which is the 'industry standard' approach).

 =however the ultimate answer to your question is: if it works for you/your
 users, then it is 'right'!

  Thanks again, Beauford

 =hope it helps you along your way to MySQL expertise,
 =dn


  - Original Message -
  From: DL Neil [EMAIL PROTECTED]
  To: Beauford.2003 [EMAIL PROTECTED]; [EMAIL PROTECTED]
  Sent: Monday, December 02, 2002 10:41 AM
  Subject: Re: Table setup question
 
 
   Beauford,
   [please reply to the list - there are always others 'lurking' who will
  learn
   from the discussion (and use of the archives is a wonderful dream...)]
  
   You are correct, there will be duplication in the AlbumSong table -
but
  not
   within one field, only within one column. Album will consist of a
unique
  ID
   (IDA) and each Title will, one assumes be unique (unless two groups
use
  the
   same name - I guess that must be possible). Similarly Songlist will
also
   consist of a unique ID (IDS) and once again, one assumes that any
  repetition
   of title will be two different songs that share the same name. These
two
  ID
   'uniqueness' rules must apply for the next to work, ie you will most
  likely
   define them as PRIMARY KEY columns.
  
   The 'translation' table, breaks the 'unique' 'rule', So if song Q
 apears
  on
   albums A, C, and E as you ask, then we will see:
  
   AlbumSong
  
   IDAIDS
   11
   21
   31
  
   Note that while 1 repeats in AlbumSong, it does not repeat (a) in
   Songlist, nor in (b) one row/field of AlbumSong, eg
  
   IDA   IDS
   1,2,31
   THE ABOVE IS TOTALLY WRONG!!!
  
   In the case of AlbumSong the table will not be able to have a PRIMARY
 KEY
   column(s), but each of the individual columns should probably be
INDEXed
  for
   read-speed.
  
   Be aware that AlbumSong contains no real 'data' from an end-user
   perspective. It is a table made up purely of 'links' or 'keys' or
'IDs'
 to
   the other two tables. The Album and Songlist tables do all the 'data'
  work,
   AlbumSong does all the 'heavy lifting' to relate Album's data to
  Songlist's,
   and/or vice-versa.
  
   Apologies if this was not clear, first time round,
   =dn
  
  
  
DL,
   
OK, that helps a lot, but I am still confused. I am still struggling
  with
how you can not have duplication in at least one of the tables. A
  diagram
here may help.  So if song Q apears on albums A, C, and E - I don't
  quite
understand how AlbumSong is going to be able to know which songs are
duplicated without duplication of ID's. (There are 500 songs and
only
 36
albums).
   
Thanks
   
AlbumSonglistAlbumSong
   
IDATITLEIDSSongIDAIDS
   
1

Re: Select, mysql_fetch_array, PHP question (sorry - typo in my original email)

2002-12-03 Thread Beauford.2003
Sorry, where it says echo $item should have read echo $line['item']; Simply
a typo, but the problem is still the same.

Hi,

I'm trying to do a search on database through a web interface using PHP and
having one small problem. Example: If I search for items that cost $5.00 and
there are three of them, the query works and displays the information, but
if there is only one item the query works but doesn't display the item. I
just get a blank page.

SELECT price, item FROM list WHERE price=$5.00;

 while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {

 echo $item;  }

If I do the same search from the MySQL command line the query will display
one or more items.

Ideas?

TIA

 
 
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 http://lists.mysql.com/   (the list archive)
 
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Re: Select, mysql_fetch_array, PHP question

2002-12-03 Thread Beauford.2003
Alex,

That's it - I have a  if (mysql_fetch_row($result)) before the while loop.
I've changed it slightly and now it works fine. The little things are always
the ones that get you.

Thanks



- Original Message -
From: Alex Pukinskis [EMAIL PROTECTED]
To: Beauford.2003 [EMAIL PROTECTED]
Sent: Tuesday, December 03, 2002 7:41 PM
Subject: Re: Select, mysql_fetch_array, PHP question


 Maybe you have an if ($line = mysql_fetch_array($result)) line
 somewhere up above?  That would cause the while loop to skip the first
 record... I usually use a do...while loop with recordsets to avoid this
 problem.

 That's my best guess, without seeing more code.

 -Alex

 On Tuesday, December 3, 2002, at 01:40  PM, Beauford.2003 wrote:

  That is what I have, a simple typo on my part.
 
  - Original Message -
  From: Alex Pukinskis [EMAIL PROTECTED]
  To: Beauford.2003 [EMAIL PROTECTED]
  Sent: Tuesday, December 03, 2002 5:04 PM
  Subject: Re: Select, mysql_fetch_array, PHP question
 
 
  To be honest, I'm not sure why what you're doing works at all.  You
  don't ever assign a value to the variable $item.  Try changing
 
  echo $item;
 
  to
 
  echo $line['item'];
 
  Is there more code involved that you didn't include?
 
  -Alex
 
  On Tuesday, December 3, 2002, at 06:38  AM, Beauford.2003 wrote:
 
  Hi,
 
  I'm trying to do a search on database through a web interface using
  PHP
  and
  having one small problem. Example: If I search for items that cost
  $5.00 and
  there are three of them, the query works and displays the information,
  but
  if there is only one item the query works but doesn't display the
  item. I
  just get a blank page.
 
  SELECT price, item FROM list WHERE price=$5.00;
 
   while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
 
   echo $item;  }
 
  If I do the same search from the MySQL command line the query will
  display
  one or more items.
 
  Ideas?
 
  TIA
 
 
 
  -
  Before posting, please check:
 http://www.mysql.com/manual.php   (the manual)
 http://lists.mysql.com/   (the list archive)
 
  To request this thread, e-mail [EMAIL PROTECTED]
  To unsubscribe, e-mail mysql-unsubscribe-
  [EMAIL PROTECTED]
  Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
 
 
 
 
 
 
 





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Re: Table setup question

2002-12-02 Thread Beauford.2003
DL,

OK, I get it now. I thought there would need to be duplication in the
AlbumSonglist DB. Just one other question though. What is the
difference/benefits of doing it this way, than the way I have it.  Currently
I use two tables - the songlist table includes 2 id fields (one that
corresponds with the album title, and one that is used for the song title).
The difference between what you have said and what I have is that I have
duplicated the actual song titles instead of a pointer to the song title
This way I have eliminated one table.

Thanks again, Beauford

- Original Message -
From: DL Neil [EMAIL PROTECTED]
To: Beauford.2003 [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Monday, December 02, 2002 10:41 AM
Subject: Re: Table setup question


 Beauford,
 [please reply to the list - there are always others 'lurking' who will
learn
 from the discussion (and use of the archives is a wonderful dream...)]

 You are correct, there will be duplication in the AlbumSong table - but
not
 within one field, only within one column. Album will consist of a unique
ID
 (IDA) and each Title will, one assumes be unique (unless two groups use
the
 same name - I guess that must be possible). Similarly Songlist will also
 consist of a unique ID (IDS) and once again, one assumes that any
repetition
 of title will be two different songs that share the same name. These two
ID
 'uniqueness' rules must apply for the next to work, ie you will most
likely
 define them as PRIMARY KEY columns.

 The 'translation' table, breaks the 'unique' 'rule', So if song Q apears
on
 albums A, C, and E as you ask, then we will see:

 AlbumSong

 IDAIDS
 11
 21
 31

 Note that while 1 repeats in AlbumSong, it does not repeat (a) in
 Songlist, nor in (b) one row/field of AlbumSong, eg

 IDA   IDS
 1,2,31
 THE ABOVE IS TOTALLY WRONG!!!

 In the case of AlbumSong the table will not be able to have a PRIMARY KEY
 column(s), but each of the individual columns should probably be INDEXed
for
 read-speed.

 Be aware that AlbumSong contains no real 'data' from an end-user
 perspective. It is a table made up purely of 'links' or 'keys' or 'IDs' to
 the other two tables. The Album and Songlist tables do all the 'data'
work,
 AlbumSong does all the 'heavy lifting' to relate Album's data to
Songlist's,
 and/or vice-versa.

 Apologies if this was not clear, first time round,
 =dn



  DL,
 
  OK, that helps a lot, but I am still confused. I am still struggling
with
  how you can not have duplication in at least one of the tables. A
diagram
  here may help.  So if song Q apears on albums A, C, and E - I don't
quite
  understand how AlbumSong is going to be able to know which songs are
  duplicated without duplication of ID's. (There are 500 songs and only 36
  albums).
 
  Thanks
 
  AlbumSonglistAlbumSong
 
  IDATITLEIDSSongIDAIDS
 
  1A   1Q??
  2B2R??
  3C3S??
  4D4T??
  5E 5V   ??
 ?
 ?
 ?
 ?
 ?
 ?
 
 
 
  - Original Message -
  From: DL Neil [EMAIL PROTECTED]
  To: Beauford.2003 [EMAIL PROTECTED]; 
  Sent: Monday, December 02, 2002 4:32 AM
  Subject: Re: Table setup question
 
 
   Hi Beauford,
  
   You are on the right track. Yes you should remove the songs to a
 separate
   table. If you merely duplicate the first example (below) in two tables
 you
   have created a one-to-many relationship between the Album table and
 the
   Songs table - one album has many songs on it. Your query code applies.
  
   However the Songs table still has duplication in it, eg 3.song appears
   thrice, and we can't have that! The problem is, if you cut down the
 Songs
   table entries so that 3.song appears only once, how do you link to
  multiple
   Albums?
  
   So now you are into a many-to-many relationship between the Album
 table
   and the Songs table - one album has many songs on it AND one song may
  appear
   on several albums. Problem!
  
   You should not have a field with 'repetition' within it, eg for each
 album
   hold two fields: ID, and Songs - where songs is a list of IDs/titles.
 This
   is not 'relational'. You can read up about a process known as
   normalisation which teaches a basic three step process (some go to
 more
   than

Table setup question

2002-12-01 Thread Beauford.2003
Hi,

I have a database of albums by a group and I want to be able to search on
this table to find out what songs are duplicated on what albums (there are
36+ albums). The problem is how do I set up the database.

Example:

Album1 has 3 songs.  1.song, 2.song, 3.song, 4.song
Album2 has 4 songs.  4.song, 5.song, 6.song, 3.song
Album3 has 4 songs.  7.song, 8.song, 1.song, 3.song

So 3.song appears on all 3 albums.

Currently I have it set up with two tables as shown below, but I am thinking
there has to be a better way to do this than to duplicate the name of the
song three, four, or five times in the table.

Table AlbumName

AlbumID
==
Album1   1
Album2   2
Album3   3

Table SongTitle

SongID
=== ==
3.song  1
3.song  2
3.song  3
7.song  3
etc.
etc.

So basically my search is  - SELECT Album, Song FROM AlbumName, SongTitle
WHERE AlbumName.ID=SongTitle.ID;

Given the setup above, is there a way that I can put in the SongTitle.ID
field that song appears on more than one album. Maybe something like:

SongID
=== ==
3.song  1, 2, 3

But then what would my search be.

Sorry for the length of this, but I am learning MySQL and trying to get a
handle on all of it. My way works, but I'm sure there has to be a better
way.

Any thoughts are appreciated.

TIA, Beauford



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Update help

2002-11-14 Thread Beauford
Hi,

I'm trying to update a table from my webpage but not having much luck,
although it works from the command line. Obviously I am missing
something and would appreciate any help.

Also. Is there a good tutorial or help page for Mysql. I find the one at
www.mysql.com not very helpful.

TIA

The code:

?

$db =mysql_connect(url,name,pass);
mysql_select_db(dbasename);
$query = UPDATE counter SET number=number+1; 

  
?



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Scrolling Databases

2002-11-12 Thread Beauford
Hi,

I have created a contact database using Mysql and want to be able to
scroll back and forth through the database. Even after I have done a
search and found a record, I want to be able to scroll back or forth
from that point on. I am using PHP4 to display the results. This may be
something easy to do, but I have not found any documentation on it, and
with my newbie skills at both Mysql and PHP, I haven't been able to put
it together. Can someone point me in the right direction. 

TIA



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RE: Errors stating Mysql server

2002-11-09 Thread Beauford
MySQL must be the owner of the data dir. Check the owner of MySQL data
dir.

It was, but I have since changed it to root and it works fine now. Any
idea why this would be?

drwxr-xr-x   12 root mysql4096 Nov  8 18:00 ./
drwxr-xr-x   15 root root 4096 Nov  8 20:12 ../
drwxr-xr-x2 root mysql4096 Nov  8 18:00 bin/
drwxr-xr-x3 root mysql4096 Nov  8 17:59 include/
drwxr-xr-x2 root mysql4096 Nov  8 17:59 info/
drwxr-xr-x3 root mysql4096 Nov  8 17:59 lib/
drwxr-xr-x2 root mysql4096 Nov  8 18:00 libexec/
drwxr-xr-x3 root mysql4096 Nov  8 18:00 man/
drwxr-xr-x6 root mysql4096 Nov  8 18:00 mysql-test/
drwxr-xr-x3 root mysql4096 Nov  8 18:00 share/
drwxr-xr-x7 root mysql4096 Nov  8 18:00 sql-bench/
drwxr-xr-x5 root mysql4096 Nov  9 05:39 var/

TIA 

-Original Message-
From: Victoria Reznichenko [mailto:victoria.reznichenko;ensita.net] 
Sent: November 9, 2002 11:28 AM
To: [EMAIL PROTECTED]
Subject: re: Errors stating Mysql server


Beauford,
Saturday, November 09, 2002, 8:27:58 AM, you wrote:

B I just reinstalled Slackware and now have the unenviable task of 
B getting Mysql working. What a major pain in the ***. Anyway, here are

B the errors. I have read manuals, searched the web, as well as the 
B newsgroups, and still can't get it to work. No matter what 
B permissions I change, the errors are still there. Is there not a 
B simple easy way to do this. This is really ridiculous.

B TIA

B 021108 21:18:11  mysqld started
B /usr/local/mysql/libexec/mysqld: File './beauford-bin.009' not found
B (Errcode: 13)
B 021108 21:18:11  Could not use beauford-bin for logging (error 13) 
B 021108 21:18:11  /usr/local/mysql/libexec/mysqld: Can't create/write 
B to file '/usr/local/mysql/var/beauford.pid' (Errcode: 13) 021108 
B 21:18:11  /usr/local/mysql/libexec/mysqld: Can't find file: 
B './mysql/host.frm' (errno: 13) 021108 21:18:11  
B /usr/local/mysql/libexec/mysqld: Error on delete of 
B '/usr/local/mysql/var/beauford.pid' (Errcode: 2) 021108 21:18:11  
B mysqld ended

MySQL must be the owner of the data dir. Check the owner of MySQL data
dir.




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Errors stating Mysql server

2002-11-08 Thread Beauford
I just reinstalled Slackware and now have the unenviable task of getting
Mysql working. What a major pain in the ***. Anyway, here are the
errors. I have read manuals, searched the web, as well as the
newsgroups, and still can't get it to work. No matter what permissions I
change, the errors are still there. Is there not a simple easy way to do
this. This is really ridiculous.

TIA

021108 21:18:11  mysqld started
/usr/local/mysql/libexec/mysqld: File './beauford-bin.009' not found
(Errcode: 13)
021108 21:18:11  Could not use beauford-bin for logging (error 13)
021108 21:18:11  /usr/local/mysql/libexec/mysqld: Can't create/write to
file '/usr/local/mysql/var/beauford.pid' (Errcode: 13)
021108 21:18:11  /usr/local/mysql/libexec/mysqld: Can't find file:
'./mysql/host.frm' (errno: 13)
021108 21:18:11  /usr/local/mysql/libexec/mysqld: Error on delete of
'/usr/local/mysql/var/beauford.pid' (Errcode: 2)
021108 21:18:11  mysqld ended



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   http://lists.mysql.com/   (the list archive)

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