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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