รบกวนดูโค้ด login ให้ผมอีกครั้งนะครับ จากที่ผมลองใช้คำสั่ง echo นอก if ดูแล้ว ปัญหาที่เกิดขึ้นคือมันไม่ยอมทำในคำสั่ง if ให้ รบกวนหน่อยครับ

<?
//echo $_POST["username"];
//session_start();
$userid=$_POST["username"];
$password=$_POST["password"];
if ($userid && $password)
{
  // if the user has just tried to log in

  $db_conn = mysql_connect("localhost", "root", "");
  mysql_select_db("project_db", $db_conn);
  $query = "select * from student "
           ."where StudentId='$userid' "
           ." and Password='$password'";
  $result = mysql_query($query, $db_conn);
  if (mysql_num_rows($result) >0 )
  {
	 //echo "ok";
    // if they are in the database register the user id
    $valid_user = $userid;
    session_register("valid_user");
  }
    if (session_is_registered("valid_user")) {
   // echo "You are logged in as: <strong>$valid_user </strong>";
	// echo "<br><br>";
	}

 } else {
    if (isset($userid))   { // if they've tried and failed to log in
  //    echo "Could not log you in";
    } else { // they have not tried to log in yet or have logged out
 //     echo "You are not logged in.<br>";
    }

    // provide form to log in
 }

?>[