我已经创建了如下所示的构造函数，但是它不起作用

Question

构造函数在laravel 6中不起作用。我已将构造函数创建为在下面给出，但不起作用

 public function __construct(Request $request){
     $status   = 0;
     $message  = "Some error occurred. Please try again later!";
     $response = (object)array();
     $json = json_decode(file_get_contents("php://input"),true);
      if($json == null || count($json) == 0 || empty($json)) {
       return response()->json(['status'  => $status,'message' => $message,'data'=> $response]);exit();
        }
    }

Answer 1

如果要定义仅处理单个动作的控制器，则可以在控制器上放置单个__invoke方法：

<?php

namespace App\Http\Controllers;

use App\Http\Controllers\Controller;

class MyController extends Controller
{
    public function __invoke(Request $request)
    {
        $status   = 0;
        $message  = "Some error occurred. Please try again later!";
        $response = (object) array();
        $json = json_decode(file_get_contents("php://input"), true);

        if ($json == null || count($json) == 0 || empty($json)) {
            return response()->json(['status' => $status, 'message' => $message, 'data' => $response]);exit();
        }
    }
}

然后为您的单个动作控制器注册一条路线：

Route::get('/test', 'MyController');