change password script

Je leest hem uit;

<?
if ($_POST['submit'] && $_POST['paswoord'])
{
$sql = "UPDATE tabel SET paswoord = '" . $_POST['paswoord'] . "' WHERE id = '" . $_POST['idnummer'] . "'";
$res = mysql_query($sql);

echo "paswoord aangepast";
}
else
{
$sql = "SELECT paswoord FROM tabel WHERE id = " . $idnummer;
$res = mysql_query($sql);
$row = mysql_fetch_array($res);

echo "<form method=\"post\" action=\"" . $_SERVER['PHP_SELF'] . "\">";
echo "paswoord:<br />";
echo "<input type=\"text\" name=\"paswoord\" value=\"" . $row['paswoord'] . "\"><br />";
echo "<input type=\"submit\" name=\"submit\" value=\"paswoord wijzigen\">";
echo "</form>";
}
?>

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/httpd/vhosts/xxxxxxx/httpdocs/changepass.php on line 34

huh?

volgens mij heb ik alles goed ingevuld...

kheb het al:

$sql = "SELECT paswoord FROM tabel WHERE id = " . $sess_id;

IPV:

$sql = "SELECT paswoord FROM tabel WHERE id = " . $idnummer;

en zo dus ook:

WHERE id = '" . $sess_id . "'";

IPV:

WHERE id = '" . $_POST['idnummer'] . "'";