PHP gevorderde |
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met deze code, krijg ik de onderstaande fout, maar snap alleen niet waarom...
Kan iemand me helpen?
<?
include "config.php";
$selectdown = mysql_query("SELECT * FROM downloads INNER JOIN koppel ON categorie.id = downloads.cat where categorie.id = '".$_GET['id']."'");
while ($make = mysql_fetch_assoc($selectdown)){
echo "".$make['id']."";
}
?>
<? include "config.php"; $selectdown = mysql_query("SELECT * FROM downloads INNER JOIN koppel ON categorie.id = downloads.cat where categorie.id = '".$_GET['id']."'"); } ?>
Citaat: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in c:program fileseasyphp1-8wwwschoolpercat.php on line 7
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