[英]Laravel 5.1: How to use middleware in an implicit controller?
我想檢查身份驗證以訪問名為FoodController的控制器,並且該控制器訪問將僅允許某些用戶,因此我創建了一個中間件並將其分配給route.php文件上的隱式控制器路由。 但這顯示了一個錯誤-
ErrorException in Router.php line 612:
strpos() expects parameter 1 to be string, array given
代碼如下:
Route.php
Route::group(['middleware' => 'auth'],function(){
Route::controller('home','HomeController');
Route::controller('food',['middleware' => 'fm', 'uses' => 'FoodController']);
});
kernel.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);
}
}
我該如何解決?
您不能將數組放入Route :: controller。 只需添加
public function __construct() {
$this->middleware('fm');
}
到您的FoodController類
解決! 您不能直接在隱式控制器中添加中間件。 請點擊以下鏈接: http : //arunkp.in/adding-middleware-in-implicit-controllers/
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.