On Saturday 02 August 2003 11:48, Lee Templeton wrote:
> The first row retrieved from the results list works and returns the row
> with Resource id #3 prefixed to the prinouts. After the first row all
> futher retrievals fail.
The variable $results is stressed by being forced to do two different
When you get that error, it generally means your query failed, or you
have the wrong variable in a mysql_fetch_*() function. Use mysql_error()
to see what the error is if the query failed.
---John W. Holmes...
PHP Architect - A monthly magazine for PHP Professionals. Get your copy
today. http://
Oops, you have one more error that I see, one one line you reference
author-cat as a table name and on another you refer to it as author_cat.
Make the appropriate change and try it again. One of the table references is
wrong and will generate a missing table error.
Jim
---Original Messag
lumn to be called or leave
the "as somenewcolname" part off if you don't care what the column name is.
This should help.
Jim
---Original Message---
From: Wilmar Perez
Date: Friday, September 06, 2002 12:16:28
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] not a vali
Hi guys
Thanks for your help. Peter was right I had an error in the sentence, I
worked it out but still the same error came up. The mysql_error() function
doesn't return any thing.
Any idea?
Thanks a lot.
>Hi
>Couple of points, you probably have an SQL error,
>I use
>echo $query;
>echo
Hi
Couple of points, you probably have an SQL error,
I use
echo $query;
echo ''.mysql_error();
which shows the query and the error
you use 'author-cat' in the table name
and 'author_cat' and 'author-cat' in the where clause which will throw an
error.
HTH
Peter
---
Hello,
It's always good practice to put some error detection in your code, i.e.
$result = mysql_query($query)
or die(mysql_error());
I think your problem is the mysql does not like your query, but you are not
catching it until PHP tries to figure out the number of rows returned by the
query.
e: [PHP-DB] Not a valid MySQL result resource
> //.../... first part of the code is to connect to the right DB on a MySQL
> server. It works fine
> //This code to show the query. It runs well under MySQL and gives 1 result
:
> $query="SELECT Login, Password FROM `user` WHER
I'd check for mysql_num_rows( $result_query) > 0 as well.
It's entirely possible the user mis-typed the login or password values.
Miles
At 07:26 PM 11/8/01 -0600, Paul DuBois wrote:
>At 2:04 AM +0100 11/9/01, MPropre wrote:
>>>//.../... first part of the code is to connect to the right DB on a My
> //.../... first part of the code is to connect to the right DB on a MySQL
> server. It works fine
> //This code to show the query. It runs well under MySQL and gives 1 result :
> $query="SELECT Login, Password FROM `user` WHERE user.login=''$login'' and
> user.password=''$password''";
> //This
At 2:04 AM +0100 11/9/01, MPropre wrote:
>//.../... first part of the code is to connect to the right DB on a MySQL
>server. It works fine
>
>
>//This code to show the query. It runs well under MySQL and gives 1 result :
> $query="SELECT Login, Password FROM `user` WHERE user.login=''$login'' and
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