Ajax JSON Decode Slim PHP问题？

好吧,我似乎无法从我的ajax调用解码JSON,它将用户数据发布到我的api,它是用超薄的php构建的.

这是我的ajax ..

var jsonData;
jsonData = [
      {
        username: "user",
        password: "pass"
      }
    ];

$.ajax({
  type: "POST",
  url: "http://localhost/api/user/auth",
  data: {
    user: JSON.stringify(jsonData)
  },
  contentType: "application/json; charset=utf-8",
  dataType: "json",
  success: function(data) {
    alert("You are good!");
  },
  error: function(xhr, type) {
    alert("Y U NO WORK?");
  }
});

这是我的SLIM PHP代码..

$app->post('/user/auth', function () use ($app) {
    try {
         $requestBody = $app->request()->getBody(); //This works

         //RequestBody is: user=%5B%7B%22username%22%3A%22user%22%2C%22password%22%3A%22pass%22%7D%5D         

         $json_a = json_decode($requestBody); //This doesn't work

         print_r($json_a); //Has no output?

         $username = $json_a['user']['username']; //Therefore this doesn't work?

    } catch(Exception $e) {
         echo '{"error":{"text": "'. $e->getMessage() .'"}}';
    }
});

正如您在注释中所看到的那样,requestBody等于：

user=%5B%7B%22username%22%3A%22user%22%2C%22password%22%3A%22pass%22%7D%5D  

但是,我似乎无法解码它,因为print_r($json_a)没有效果.

非常感谢任何帮助,谢谢.

解决方法:

尝试

$params_str = urldecode($requestBody);
parse_str($params_str, $params_arr);
$user = json_decode($params_arr['user']);

标签：zepto,javascript,php,json,ajax
来源： https://codeday.me/bug/20190901/1784620.html