Here is my AJAX code:
 $("#loginbutton").click(function(){
              var email = $('#email').val();
              var password = $('#password').val();
              $.ajax({
               url: 'login.php',
               type: 'POST',
               dataType: 'json',
               data: { email:email, password:password},
               success: function(data){
                 $("#samplediv").innerHTML ="Welcome";
                }
              });
             });
And this is my PHP:
 <?php
 session_start();
 $conn = mysqli_connect("localhost", "root", "", "getpet");
 if(isset($_POST['email'])){
   echo $_POST['email'];
 }
?>
As you can see it will be a login system, but the php doesn't write out the $_POST['email']
variable. I think it is probably a syntax mistake, just I am too blind. I would really appreciate it if somebody can  help me.
UPDATE: Here is my whole php code, i think it's not relevant, but this is ther reason, why i use dataType: json.
 <?php
 session_start();
 $conn = mysqli_connect("localhost", "root", "", "getpet");
 if(isset($_POST['email'])){
   echo $_POST['email'];
 }/*
 $results = array(
   'success' => false,
   'user_id' => "azaz",
   'fname' => "",
   'lname' => ""
 );
 if(!empty($_POST['email']) && !empty($_POST['password'])){
        $email = $_POST['email'];
        $password = md5($_POST['password']);
        $sql= "SELECT * FROM users WHERE email = '$email' AND password = '$password'";
        $rowsql = mysqli_query($conn, $sql);
        $row = mysqli_fetch_array($rowsql, MYSQLI_BOTH);
        if(mysqli_num_rows($rowsql) == "1"){
            $_SESSION['user_id'] = $row['user_id'];
            $_SESSION['fname'] = $row['fname'];
            $_SESSION['lname'] = $row['lname'];
      $results['success'] = true;
      $results['user_id'] = $row['user_id'];
      $results['fname'] = $row['fname'];
      $results['lname'] = $row['lname'];
      }
    }
            //echo json_encode($results);*/
 ?>
 
     
     
     
    