Codice PHP:
$query2 = "SELECT * FROM option WHERE Select = '$IDCampo' AND Selected = '1'";
$result2 = mysql_query($query2, $db);
echo mysql_error($db);
$row2 = mysql_fetch_array($result2);
$ValueOption = $row2["Value"]; 
Perché mi da':

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/local/psa/home/vhosts/sito.it/httpdocs/pages/struttura.php on line 36