为什么我不能使用Ajax和Jquery将数据作为json传递到php端？

Question

function delete_user(email)
{

    //if(confirm("Do you want to remove this user with mail id: "+email))
    //{ 
        var z = {email : email}

        console.log(z)
        $.ajax({
        url: "delete_user.php",
        type: 'POST',
        data: z,

        success: function (data) {


            console.log(data);
            return false;
            var i=0;
            var element = "<table style='border-collapse: collapse;'>"
            while(data[i] != null)
            {

                element = element + "<tr style='height:15px' onmouseover='rowbig(this)' onmouseout='rowsmall(this)'><td>"+data[i].fname+"&nbsp"+data[i].lname+"</td><td>"+data[i].email+"</td><td>"+data[i].dob+"</td><td><button onclick='delete_user(\""+data[i].email+"\")'><img src='delete.png' style='height:20px;width:20px'/></button></td></tr>";               
                i++;
            }
            element = element + "</table>";


            $("#response").html( element );



        },
        cache: false,
        contentType: false,
        processData: false
    });
    //}

}


<?php 

    header("Content-Type:application/json");
         $email ="";

         function test_input($data) {
          $data = trim($data);
          $data = stripslashes($data);
          $data = htmlspecialchars($data);
          return $data;
        }
        function deliver_res($status_message,$result)
             {
                header("$status_message");


                $response['status_message']=$status_message;
                $response['result']=$result;

                $json_response=json_encode ($response);
                echo $json_response;
             }


             $servername = 'localhost';
             $username = 'root';
             $password = '';
             $dbname = 'webdata';
             $conn = mysqli_connect($servername, $username, $password, $dbname);

            if(isset($_POST['email']))
            {
                $email = test_input($_POST["email"]);

            }
            else
            {

                deliver_res("name not received","false");
                exit();

            }


            $sqll="DELETE FROM admindata WHERE email='".$email."';";
            $result = mysqli_query($conn,$sqll);


             $return_arr = array();

            $sqll="SELECT * FROM admindata ORDER BY id DESC ;";
            $result = mysqli_query($conn,$sqll);


            while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) 
            {
                $row_array['fname'] = $row['fname'];
                $row_array['lname'] = $row['lname'];
                $row_array['email'] = $row['email'];
                $row_array['dob'] = $row['dob'];

                array_push($return_arr,$row_array);
            }

            echo json_encode($return_arr);

            echo "\n";

    mysqli_close($conn);






?>

我似乎无法将使用AJAX和Jquery的数据作为JSON传递给PHP端。语法有什么问题吗？