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Posted by Simon Bridgewater on 06/26/05 09:34
Have you tried executing this sql statement from mysqladmin or similar. Just
to see if you get a result .
"Kees Boer" <keesboer@integrity-computing.net> wrote in message
news:Hdrve.68963$gc6.64748@okepread04...
> Hi, I've got a little bug, which I'm having a hard time debugging. It's
> probably in the syntax. This is the code:
>
> <?php
>
> $var_film = 'filmbigfish';
>
> // connect to mysql_server
> $mysqllink = mysql_connect( "localhost", "userid", "somepw" );
> // select database to use
> mysql_select_db( "kboer" );
>
> // primary table that directory uses
> $table = "films";
> $qresult = mysql_query( "SELECT * FROM `films` WHERE `d_thisfilm` =
> `$var_film` " );
>
> $row = mysql_fetch_assoc($qresult);
>
> ?>
>
> <html>
> <head>
> <title>Untitled Document</title>
> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
> </head>
>
> <body bgcolor="#FFFFFF" text="#000000">
>
> Everything works!
>
> </body>
> </html>
>
> I'm getting the following error.
>
> Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL
result
> resource in
>
/home/kboer/positive-entertainment.com/public_html/films/pages/resultlink.ph
p
> on line 14
>
> What is the problem?
>
> Thanks!
>
> Kees
>
>
>
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