Ik heb dit script:
<?php
error_reporting(E_ALL);
include("config.php");
$query = "SELECT le.naam, le.site, le_categorie.categorie".
"FROM le,le_categorie ".
"WHERE le.categorie = le_categorie.categorie";
$result = mysql_query($query);
$content = array();
while($row = mysql_fetch_array($result)){
$content[$row['categorie']][] = $row['naam'];
}
echo "<table width=\"100%\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\" align=\"left\">
<tr>
<td width=\"285\" height=\"14\" bgcolor=\"#096ea1\">
<span class=\"style1\">Links:</span></td>
</tr>
</table>";
foreach($content as $categorie => $namen)
{
echo "<b><br />" . $categorie. ":</b><br />";
foreach($namen as $naam)
{
echo'
<body link="#000000" vlink="#000000" alink="#000000">
<a href="'.$site.'">'.$naam.'</a><br />
</body>';
}
}
?>
Het geeft deze melding:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/httpd/vhosts/duursportersweb.nl/httpdocs/links/overzicht.php on line 11
Wat doe ik fout?
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