io ho fatto come hai detto tu ovvero:
codice:
 
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd" >
<html>
<head>
<script language=javascript>
	function elimina(codice)
	 { 
       f.cod.value=codice;
       f.action="del_usr.php";
       f.submit();
     }
</script>

function mostra_lista()
{
	if(isset($_GET['msg']))
		echo ''.htmlentities($_GET['msg']).'

';

	$query = "SELECT id, uid, upwd FROM members";
	$result = mysql_query($query);
	if (!$result) {
		die("Errore nella query $query: " . mysql_error());
	}
	echo '
	<div align=center valign=center><table border="1">
		<tr>
			<th>id</th>
			<th>utente</th>
			<th>password</th>
			<th></th>
		</tr>';
 	while ($row = mysql_fetch_assoc($result))
	{
		$id=htmlspecialchars($row['id']);
		$uid=htmlspecialchars($row['uid']);
		$upwd=htmlspecialchars($row['upwd']);
		$link = $_SERVER['PHP_SELF'] . '?id=' . $row['id'];
 
		echo "<form name=f method=post><tr>
			  <input type=hidden name=cod>
				<td>$id</td>
				<td>$uid</td>
				<td>$upwd</td>
				<td><a href=\"$link\">modifica</a></td>
				<td><input type=button value=elimina onclick=elimina('$row[0]')></input></td>
			</tr></form>";
	}
	echo '</table></div>';
	mysql_free_result($result);
	mysql_close();
}
Ma niente da sempre il solito errore