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:
>><?php
>>//.../... 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 var should contain a query result ressource:
>> $result_query=mysql_query($query);
>>
>>//Here's my error message after executing:
>
>Not so fast. Where's your error checking to verify that the
>query actually succeeded?
>
>if (!$result_query)
>{
> die ("Gee, I guess error checking is a good thing after all!"
> . mysql_error ());
>}
>
>>// Supplied argument is not a valid MySQL result resource in this line:
>> $result_table=mysql_fetch_array($result_query);
>>
>>// I thought that this var ($result_table) should contain the row of 2 cells
>>//And I expected to read $result_table['login'] or $result_table[0] as the
>>first cell of the row...
>>// I any of you can help me, poor php newby... :o) Great thanks !!
>>
>>?>
>>
>>
>>
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