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Home/ Questions/Q 960371
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T01:10:22+00:00 2026-05-16T01:10:22+00:00

Hey there, why does this code not work? $qry = mysql_query(SELECT performerid,pic0 FROM .$table.

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Hey there, why does this code not work?

$qry = mysql_query("SELECT performerid,pic0 FROM ".$table." ORDER BY RAND() LIMIT 6"); 

        $start =  new WP_Query('showposts=6&orderby=rand');

        if ($start->have_posts()) : while( $start->have_posts() ) : $start->the_post();

        $rows = mysql_fetch_assoc($qry);

        if (!$rows) 
           {
           mysql_data_seek($rows,0);
           $rows = mysql_fetch_assoc($qry);
           }

        $perfs = $rows['performerid'];

        $pics  = $rows['pic0']; 

I ahve the following error:

Warning: mysql_data_seek(): supplied argument is not a valid MySQL result resource in /home/content/d/d/a/ddxxxx
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T01:10:23+00:00Added an answer on May 16, 2026 at 1:10 am

    Your call to mysql_data_seek only happens if $rows is null. If that’s true, then the call to mysql_data_seek will certainly fail, because one of it’s required args is null. That’s why you’re getting the error message.

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