Ajax请求post方式

得到post.php返回的数据
前端代码如下：

<!DOCTYPE html>
<html lang="en">
<head>
  <meta charset="UTF-8">
  <title>POST</title>
</head>
<body>
<script> var xhr = new XMLHttpRequest(); //注意路径 xhr.open("POST", "http://localhost/request/POST.php"); xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded"); xhr.send("name=lc&id=1"); xhr.onreadystatechange = function () {
     if (xhr.status == 200 && xhr.readyState == 4) {
     console.log(JSON.parse(xhr.responseText)); } } </script>
</body>
</html>

本地POST.php

<?php
$name = $_POST['name'];
$id = $_POST['id'];
echo json_encode(array(
    'name' => $name,
    'id' => $id,
));

ok 完事