wenn ich den code nehme:
dann kommt immer der fehler
why das???
PHP-Code:
<font face=Verdana size=1><?
include ('config.inc.php');
$conection = mysql_connect ($host,$usr,$pw);
mysql_select_db($db,$conection);
$SQLstring1 = "SELECT * FROM `user` by rang DESC";
$ergebnis1 = mysql_query($SQLstring1,$conection);
echo '<table bgcolor=#6B7D9A style="BORDER-COLLAPSE: collapse" borderColor=#111111 cellSpacing=0 cellPadding=0 width="100%" border=1><tr><th>Membername</th><th>Id</th><th>E-Mail</th><th>ICQ</th><th>Section</th><th>Status</th><th>Rang</th></tr>';
while($zeile=mysql_fetch_row($ergebnis1))
echo '<tr><td width="33%" bgcolor=#6B7D9A align=center>'.$zeile[0].'</td><td width="33%" bgcolor=#6B7D9A align=center>'.$zeile[3].'</td><td width="33%" bgcolor=#6B7D9A align=center>'.$zeile[2].'</td></tr>';
echo '</table>';
?>
</font>
Code:
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/www/n4b12/html/billard-liga/rsn/member.php on line 8
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