php - Laravel 5.1：如何在隐式控制器中使用中间件？

Question

我想检查身份验证以访问名为 FoodController 的控制器，并且此控制器访问权限仅允许某些用户，因此我创建了一个中间件并将其分配给 route.php 文件上的隐式控制器路由。但它显示一个错误 -

ErrorException in Router.php line 612:
strpos() expects parameter 1 to be string, array given

代码如下：

路由.php

Route::group(['middleware' => 'auth'],function(){
    Route::controller('home','HomeController');
    Route::controller('food',['middleware' => 'fm', 'uses' => 'FoodController']);
});

内核.php

protected $routeMiddleware = [
    'auth' => \App\Http\Middleware\Authenticate::class,
    'auth.basic' => \Illuminate\Auth\Middleware\AuthenticateWithBasicAuth::class,
    'guest' => \App\Http\Middleware\RedirectIfAuthenticated::class,
    'fm' => \App\Http\Middleware\FoodAuthentication::class,
];

FoodAuthentication.php

<?php

namespace App\Http\Middleware;

use Closure;
use Illuminate\Support\Facades\Auth;

class FoodAuthentication
{
    /**
     * Handle an incoming request.
     *
     * @param  \Illuminate\Http\Request  $request
     * @param  \Closure  $next
     * @return mixed
     */
    public function handle($request, Closure $next)
    {
        if (Auth::user()->role!=2) {
            return redirect('home');
        }
        return $next($request);
    }
}

我该如何解决？

Answer 1

您不能将数组放入 Route::controller。只需添加

public function __construct() {
       $this->middleware('fm');
  }

到你的 FoodController 类

Answer 2

解决！您不能直接在隐式控制器中添加中间件。按照以下链接： http ://arunkp.in/adding-middleware-in-implicit-controllers/