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