PHP实现简单接口

ajax请求后端接口返回数据是如何实现的呢
<?php

$name = $email = '';

function test_input($data) {
    if(isset($data) && !empty($data))
    {
        $data = trim($data);
        $data = stripcslashes($data);
        $data = htmlspecialchars($data);
        return $data;
    }
}

if($_SERVER["REQUEST_METHOD"] == "POST")
{
    if(isset($_POST["name"]) && !empty($_POST["name"]))
    {
        $name = test_input($_POST["name"]);

    }
    if(isset($_POST["email"]) && !empty($_POST["email"]))
    {
        $email = test_input($_POST["email"]);
    }
}

$data = array(
    'code' => 1,
    'data' => array(  // 数组包含对象
        'list' => array(
            'name' => $name,
            'email' => $email
        ),
        'name' => $name . 'Yang'
    ),
    "content" =>[$name,$name,$name], // 数组
    'message' => 'success'
);

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转载自blog.csdn.net/nick_yangxiaotong/article/details/80890938