Laravel 尝试将一条路由分配给多个中间件

为多个中间件分配一个路由时出现问题 这是我在web.php中看到的

    Route::group(['middleware' => ['hodAndProvost']], function () {
    Route::match(['post','get'],'applied', 'LeavesController@applied');
    Route::match(['post','get'],'approval/{id}', 'LeavesController@approval');
});

这是我的App\htpp\kanel.php中的内容

protected $middlewareGroups = [
    'web' => [
        \App\Http\Middleware\EncryptCookies::class,
        \Illuminate\Cookie\Middleware\AddQueuedCookiesToResponse::class,
        \Illuminate\Session\Middleware\StartSession::class,
        // \Illuminate\Session\Middleware\AuthenticateSession::class,
        \Illuminate\View\Middleware\ShareErrorsFromSession::class,
        \App\Http\Middleware\VerifyCsrfToken::class,
        \Illuminate\Routing\Middleware\SubstituteBindings::class,
    ],

    'hodAndProvost' => [
        \App\Http\Middleware\hodMiddleware::class,
        \App\Http\Middleware\provostMiddleware::class,
    ],

    'application' => [
        \App\Http\Middleware\hodMiddleware::class,
        \App\Http\Middleware\provostMiddleware::class,
        \App\Http\Middleware\lecturerMiddleware::class,
    ],

    'api' => [
        'throttle:60,1',
        'bindings',
    ],
];

这就是我的Hodmidware里的东西

public function handle($request, Closure $next)
{
    if ($request->user() && $request->user()->user_access_id != '2')
{
    return new Response(view('unauthorized')->with('role', 'HOD'));
}
    return $next($request);
}

public function handle($request, Closure $next)
{
    if ($request->user() && $request->user()->user_access_id != '3')
{
    return new Response(view('unauthorized')->with('role', 'PROVOST'));
}
    return $next($request);
}

这就是我的教务长办公室

public function handle($request, Closure $next)
{
    if ($request->user() && $request->user()->user_access_id != '2')
{
    return new Response(view('unauthorized')->with('role', 'HOD'));
}
    return $next($request);
}

public function handle($request, Closure $next)
{
    if ($request->user() && $request->user()->user_access_id != '3')
{
    return new Response(view('unauthorized')->with('role', 'PROVOST'));
}
    return $next($request);
}

这就是我得到的错误

Symfony\Component\Debug\Exception\FatalThroTableError

---

找不到类“App\Http\Middleware\Response”

您没有导入

Response

类，因此它在同一命名空间中查找它。您甚至不需要创建新的

响应

，

view（）

也是一个响应

请尝试以下代码：

<?php

namespace App\Http\Middleware;

use Closure;

class hodMiddleware
{
    public function handle($request, Closure $next)
    {
        if ($request->user() && $request->user()->user_access_id != 2) {
            return view('unauthorized')->with('role', 'HOD'); // like your example
            // OR: //
            return redirect()->back();
        }

        return $next($request);
    }
}

---

添加
使用light\Support\Facades\Response在中间件的顶部，或者您可以使用
response（）
helper.@porloscerosψ仍然不工作，还是相同的错误？尝试
returnresponse（）->view（'unauthorized'，['role'=>'PROVOST'，401）
@porloscerrosψsir问题在于不应该读取该行返回，因为该值是正确的，读取该行意味着登录用户不是教务长，或者HODI意味着您将
返回新响应（视图（'unauthorized'）->替换为（'role'，'provost'）通过
返回响应（）->查看（'unauthorized'，['role'=>'PROVOST'，401）