hallo, ich habe folgenen code: ( er soll ein Aufklappmenu darstellen)
PHP-Code:
<?
echo "[b]Navigation[/b]
";
$open = $HTTP_GET_VARS["open"];
if($open == ""){$open = "nothing";}
include('../forum/mysql.php');
$abfrage = "SELECT * FROM menu WHERE `untermenu` LIKE '$leer'";
$ergebnis = mysql_query($abfrage);
while($row = mysql_fetch_object($ergebnis))
{
echo "[b]<a href=\"show.php?open=$row->name&include=$include\">$row->name</a>[/b]
";
$abfrage2 = "SELECT * FROM menu WHERE `untermenu` LIKE '$open' AND ´untermenu´ LIKE '$row->name'";
$ergebnis2 = mysql_query($abfrage2);
while($row2 = mysql_fetch_object($ergebnis2))
{
echo "-<a href=\"$row2->url&include=$include\">$row2->name</a>
";
}
}
?>
fehler:
Code:
Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource
datenbankverbindung und etc. steht.
fehler liegt in zeile 13:
PHP-Code:
$abfrage2 = "SELECT * FROM menu WHERE `untermenu` LIKE '$open' AND ´untermenu´ LIKE '$row->name'";