simple return data jquery

1$(document).ready(function(){
2    $.ajax({
3        url: 'ajaxfile.php',
4        type: 'get',
5        dataType: 'JSON',
6        success: function(response){
7            var len = response.length;
8            for(var i=0; i<len; i++){
9                var id = response[i].id;
10                var username = response[i].username;
11                var name = response[i].name;
12                var email = response[i].email;
13
14                var tr_str = "<tr>" +
15                    "<td align='center'>" + (i+1) + "</td>" +
16                    "<td align='center'>" + username + "</td>" +
17                    "<td align='center'>" + name + "</td>" +
18                    "<td align='center'>" + email + "</td>" +
19                    "</tr>";
20
21                $("#userTable tbody").append(tr_str);
22            }
23
24        }
25    });
26});

1<?php
2
3include "config.php";
4
5$return_arr = array();
6
7$query = "SELECT * FROM users ORDER BY NAME";
8
9$result = mysqli_query($con,$query);
10
11while($row = mysqli_fetch_array($result)){
12    $id = $row['id'];
13    $username = $row['username'];
14    $name = $row['name'];
15    $email = $row['email'];
16
17    $return_arr[] = array("id" => $id,
18                    "username" => $username,
19                    "name" => $name,
20                    "email" => $email);
21}
22
23// Encoding array in JSON format
24echo json_encode($return_arr);