facendola cosė quindi?
Codice PHP:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Documento senza titolo</title>
</head>
<body>
<p>
<?
//connessione db
session_start();
$link = mysql_connect('localhost', '', '')
or die('impossibile connettersi al server: ' . mysql_error());
mysql_select_db('my_manulazph') or die('impossibile connettersi al db');
$utente = mysql_real_escape_string($_POST['utente']);
$pwd=mysql_real_escape_string($_POST['pwd']);
$query = "SELECT * FROM utenti WHERE ID='$utente' && PASSWORD='$pwd'";
$result = mysql_query($query) or die('Query fallita: ' . mysql_error());
$num_rows = mysql_num_rows($result);
echo $result;
$utenti= $result ;
?>
</p>
<p> </p>
<table width="50%" border="0" align="center">
<tr>
<td width="24%"><form action="../interrogazione_db.php" method="post" name="inviadati" id="inviadati">
<table width="30%" "border="0" align="center" class="tabella">
<tr>
<td align="center" valign="middle" class="categoria"><strong class="descrizione_campi">UTENTE:</strong></td>
</tr>
<tr>
<td align="center" valign="middle"><p>
<label for="utente"></label>
<select name="utente" id="utente">
<option value="<?=$utenti?>"><?=$utenti?></option>
</select>
</p></td>
</tr>
<tr>
<td align="center" valign="middle" class="categoria"><strong class="descrizione_campi">PASSWORD:</strong></td>
</tr>
<tr>
<td align="center" valign="middle"><input name="pwd" type="password" class="campi" /></td>
</tr>
<tr>
<td align="center" valign="middle"><input name="inviate" type="submit" class="bottoni" value="LOGIN!" onclick=""/></td>
</tr>
</table>
</form></td>
</tr>
</table>
<p> </p>
</body>
</html>
COSė FACENDO MI STAMPA A VIDEO E DENTRO IIL MENU' A TENDINA IL SEGUENTE ERRORE: Resource id #3