如何使PHP API将JSON返回到Javascript HTTP POST/GET请求？

我一直在遵循本教程创建一个“api偏心web应用程序”，但是该应用程序只包括从PHP发出请求

在教程进行到一半时，您可以通过浏览器中的url发出请求，例如

以下是接收请求的前端控制器的代码

<?php
// Define path to data folder
define('DATA_PATH', realpath(dirname(__FILE__).'/data'));

//include our models
include_once 'models/TodoItem.php';

//wrap the whole thing in a try-catch block to catch any wayward exceptions!
try {
    //get all of the parameters in the POST/GET request
    $params = $_REQUEST;

    //get the controller and format it correctly so the first
    //letter is always capitalized
    $controller = ucfirst(strtolower($params['controller']));

    //get the action and format it correctly so all the
    //letters are not capitalized, and append 'Action'
    $action = strtolower($params['action']).'Action';

    //check if the controller exists. if not, throw an exception
    if( file_exists("controllers/{$controller}.php") ) {
        include_once "controllers/{$controller}.php";
    } else {
        throw new Exception('Controller is invalid.');
    }

    //create a new instance of the controller, and pass
    //it the parameters from the request
    $controller = new $controller($params);

    //check if the action exists in the controller. if not, throw an exception.
    if( method_exists($controller, $action) === false ) {
        throw new Exception('Action is invalid.');
    }

    //execute the action
    $result['data'] = $controller->$action();
    $result['success'] = true;

} catch( Exception $e ) {
    //catch any exceptions and report the problem
    $result = array();
    $result['success'] = false;
    $result['errormsg'] = $e->getMessage();
}

//echo the result of the API call
echo json_encode($result);
exit();

---

要调用API，需要使用JavaScript发出AJAX请求。请阅读。本页有分步指南

---

当您从API发送JSON时，在AJAX成功后，您可能需要从API发送接收到的内容。

下面是一个简单的示例

<?php
session_start();
if(!isset($_SESSION['user'])) {
    echo -1;
    die;
}
$email=$_SESSION['user'];
$arr= array();
$con=mysqli_connect("localhost","usrname","password","databasename");
if (mysqli_connect_errno()) 
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="select * from user where email = '$email'";
$result = mysqli_query($con,$sql);

while ( $row = mysqli_fetch_array($result)) {
        $arr[] = $row['u_id'];
        $arr[] = $row['f_name'];
        $arr[] = $row['l_name'];
        $arr[] = $row['email'];
        $arr[] = $row['telephone'];
        $arr[] = $row['address'];
}
echo json_encode($arr);
mysqli_close($con);?>

这里的响应包含来自数据库的信息。URL参数是php页面的URL。

---

我希望这能对您有所帮助。

所以我让它开始工作，我尝试的是Javascript中的常规XMLHttpRequest。进程正在工作，因为数据正在保存，但没有返回任何内容。我缺少的是index.php顶部的这一行

header('Access-Control-Allow-Origin: *');

---

我理解*意味着所有网站都可以访问API

看看php中的json_编码。这是@georoot的链接，代码已经使用了json_编码。问题似乎是关于如何从JavaScript发出适当的请求。很抱歉之前没有提到这一点，但请求将跨域发出，我认为这种方式不起作用，对吗？

header('Access-Control-Allow-Origin: *');