无法使用PHP JSON连接到MySql

Question

我有以下PHP代码处理由JSON

发布的某些表单AJAX对象

我的问题是没有任何东西被添加到MySql表中。因此，我一直将referenceNumber视为空。

有什么建议吗？

PHP

<?php

if (is_ajax()) {
  if (isset($_POST["action"]) && !empty($_POST["action"])) { //Checks if action value exists
    $action = $_POST["action"];
    switch($action) { //Switch case for value of action
      case "test": test_function(); break;
    }
  }
}

//Function to check if the request is an AJAX request
function is_ajax() {
  return isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest';
}

function test_function(){

    $firstName  = $_POST["firstName"];
    $lastName   = $_POST['lastName'];
    $name       = $firstName. ' ' .$lastName ;
    $email      = $_POST['email'];
    $phone      = $_POST['phone'];
    $subject    = $_POST['subject'];
    $hear       = $_POST['hear'];
    $message    = $_POST['message'];

    date_default_timezone_set("EST");
    $date = date("Y.m.d"); // 2015.10.01
    $day = date("l"); // Sunday
    $time = date("g:ia"); // 1:25am

    // Create connection
    $conn = new mysqli("localhost", "aaa", "xxx", "onlineMessages");
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }

    $add = mysqli_query("INSERT INTO `test` (`Date`, `Day`, `Time`, `Name`, `Email`, `Phone`, `Subject`, `From`, `Message`)
        VALUES ('$date','$day','$time','$name','$email','$phone','$subject','$hear','$message')");


    $get_ref_number = mysqli_query(" SELECT `Reference #` FROM `test` where Name='$name' and Phone='$phone' and Message='$message' ");
    $ref_number_fetch = mysqli_fetch_array($get_ref_number);
    $ref_number = $ref_number_fetch['Reference #'];

    // Close connection
    mysqli_close($link);

    //Returned Data
    $arr = array('Name' => $name, 'Phone' => $phone, 'referenceNumber' => $ref_number);
    echo json_encode($arr);

}
?>

JS

$(document).ready(function(){

      var form = $('#contact-form');

      form.submit(function(){
        var data = {
          "action": "test"
        };
        data = form.serialize() + "&" + $.param(data);
        $.ajax({
              type: "POST",
              dataType: "json",
              url: form.attr('action'),
              data: data,
              success: function(data) {
                  $(".the-return").html(
                    data["Name"] + data["Phone"] + data["referenceNumber"]
                  );
                  alert("Form submitted successfully.\nReturned json: " + data["json"]);
              },
              error : function(){
                  form.html("Error!" + data["json"]);
              }
        });
        return false;
      });
});