Bl4dE
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Code:
<?
//verbinden
$db = mysql_connect("localhost","user","pw");
//auslesen
$SQL = "SELECT * FROM linktit ORDER BY LinkID";
$resultat = mysql_db_query("ctchen",$SQL);
while($row = mysql_fetch_array($resultat)) {
extract($row);
echo '<table width="99%" border="0">
<tr>
<td width="33%">'.$name.'</td>
<td width="33%">'.$info.'</td>
<td width="33%">'.$link.'</td>
</tr>
</table>
';
}
echo "<br>";
mysql_close($db);
?>
Code:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/export/www/vhosts/funnetwork/hosting/ctchen/v.2.0/links.php on line 23
das is die while()-schleife?! woran liegt das?
danke im vorraus
Zuletzt bearbeitet: