如何使用AJAX将数据POST到php文件

我在将数据发送到要处理的php文件时遇到问题。我几乎什么都试过了，但找不到问题的根源。下面是一个php文件，在用户单击购买按钮后，该文件将产品名称、价格和ID发送到checkout函数。

<?php
      $servername = "localhost";
      $username = "root";
      $password = "root";
      $dbname = "Test";
        // Create connection
      $conn = new mysqli($servername, $username, $password, $dbname);
        // Check connection
      if ($conn->connect_error) {
       die("Connection failed: " . $conn->connect_error);
     } 
     $sql = "SELECT P1.Product_Name, S2.Price, P1.Product_ID FROM Product P1, Sale_Item S2 WHERE P1.Product_ID=S2.Product_ID AND P1.Category='Sports'";
     $res = $conn->query($sql);
     $counter=0;
     while ($row = $res->fetch_assoc()){
      $Product_Name = $row["Product_Name"];
      $Price = $row["Price"];
      $Product_ID = $row["Product_ID"];
      echo ('<td><p></p>'.$row["Product_Name"].'<br>'.$row["Price"].'<p></p><input type="button" value="Buy" onclick="checkout(''' . $Product_Name . ''', ''' . $Price . ''', ''' . $Product_ID . ''')"</td>');          
      $counter++;
      if ($counter==3) {
        $counter=0;
        print "<br>";
      }
    }
    $conn->close();
    ?>

接下来是checkout函数：

<script type="text/javascript">
        function checkout(Product_Name, Price, Product_ID) {
          //document.write(Product_Name, Price, Product_ID)
          var theProduct_Name = Product_Name;
          var thePrice = Price;
          var theProduct_ID = Product_ID;
          $.ajax({
            type: "POST",
            url: "http://localhost:8888/checkout.php",
            data: {Product_Name: theProduct_Name, Price: thePrice, Product_ID: theProduct_ID},
          });
          window.location.assign("http://localhost:8888/checkout.php")
        }
      </script>

我使用的是MAMP的phpMyAdmin数据库。我的url不正确吗？我试过使用"http://localhost:8888/checkout.php"，而只使用checkout.php。下面是我需要处理数据的php文件。为了简单地学习如何发送数据，我只是在文件中回显，以确保它确实在发布。但没有任何回应。

<?php 
  session_start();
  $servername = "localhost";
  $username = "root";
  $password = "root";
  $dbname = "Test";
  // Create connection
  $conn = new mysqli($servername, $username, $password, $dbname);
  // Check connection
  if ($conn->connect_error) {
   die("Connection failed: " . $conn->connect_error);
 } 
  $theProduct_Name = $_POST['Product_Name'];
  $theProduct_ID = $_POST['Product_ID'];
  $thePrice = $_POST['Price'];
 echo $theProduct_Name.$theProduct_ID.$thePrice;
 ?> 

我是网络编程的新手，所以任何帮助或提示都将不胜感激。我已经看了几个小时了，似乎无法让它发挥作用。