将中间件应用于除 setup/* 之外的所有路由 Laravel 5.4

Question

我正在 Laravel 应用程序中试验中间件。我目前已将它设置为运行在每个经过身份验证的用户的路由上，但是，我希望它忽略任何以 setup URI.

开头的请求

这是我的 CheckOnboarding 中间件方法的样子：

public function handle($request, Closure $next)
{
    /** 
    * Check to see if the user has completed the onboarding, if not redirect.
    * Also checks that the requested URI isn't the setup route to ensure there isn't a redirect loop.
    */
    if ($request->user()->onboarding_complete == false && $request->path() != 'setup') {
        return redirect('setup');
    } else {
        return $next($request);
    }
}

这在我的路线中是这样使用的：

Route::group(['middleware' => ['auth','checkOnboarding']], function () {
    Route::get('/home', 'HomeController@index');
    Route::get('/account', 'AccountController@index');

    Route::group(['prefix' => 'setup'], function () {
        Route::get('/', 'OnboardingController@index')->name('setup');
        Route::post('/settings', 'SettingsController@store');
    }); 
});

现在，如果我转到 /home 或 /account，我会像您期望的那样被重定向到 /setup。这最初导致了重定向循环错误，因此 & $request->path() != 'setup' 在中间件中。

我觉得这是一种非常笨拙的方法，并且在 setup 之后显然不匹配任何内容，就像我创建的 setup/settings 路由一样。

有没有更好的方法让这个中间件运行出现在用户的所有路由上，同时设置某些路由应该免于此检查？

Answer 1

不希望中间件到达的路由运行，只需将它们放在函数之外即可：

//here register routes on which you dont want the middleware: checkOnboarding
Route::group(['middleware' => ['auth','checkOnboarding']], function () {
     //routes on which you want the middleware
});

Answer 2

您所做的没有任何问题，但是，我建议您将路线组分开，即：

Route::group(['middleware' => ['auth', 'checkOnboarding']], function () {
    Route::get('/home', 'HomeController@index');
    Route::get('/account', 'AccountController@index');
});

Route::group(['prefix' => 'setup', 'middleware' => 'auth'], function () {
    Route::get('/', 'OnboardingController@index')->name('setup');
    Route::post('/settings', 'SettingsController@store');
});

或者，为您的身份验证创建一个父组：

Route::group(['middleware' => 'auth'], function () {

    Route::group(['middleware' => 'checkOnboarding'], function () {
        Route::get('/home', 'HomeController@index');
        Route::get('/account', 'AccountController@index');
    });

    Route::group(['prefix' => 'setup'], function () {
        Route::get('/', 'OnboardingController@index')->name('setup');
        Route::post('/settings', 'SettingsController@store');
    });
});

这也意味着您可以删除中间件中的额外条件：

/**
 * Check to see if the user has completed the onboarding, if not redirect.
 * Also checks that the requested URI isn't the setup route to ensure there isn't a redirect loop.
 */
return $request->user()->onboarding_complete ? $next($request) : redirect('setup');

希望对您有所帮助！

Answer 3

您可以为此使用控制器 class，并获得非常惊人的效果。

如果您在 HTTP/Controllers/Controller.php 中创建一个 __construct 函数，那么您可以在每个控制器操作上向运行声明中间件，甚至可以根据需要声明异常。

class Controller extends BaseController
{
  use AuthorizesRequests, DispatchesJobs, ValidatesRequests;
  public function __construct(){
    $this->middleware('auth',['except' => ['login','setup','setupSomethingElse']]);
  }
}

注意不要将任何标准索引、存储、更新、销毁函数置于异常中，否则您将打开潜在的安全问题。

Answer 4

有两种方法可以解决这个问题

尝试在路线文件中筛选路线 web.php or api.php
跳过 middleware

如果是全局中间件（你想在所有路由之前运行的中间件），你应该在中间件中跳过路由。

例如：

//add an array of routes to skip santize check
protected $openRoutes = [
    'setup/*',
];

/**
 * Handle an incoming request.
 *
 * @param  \Illuminate\Http\Request  $request
 * @param  \Closure  $next
 * @return mixed
 */
public function handle($request, Closure $next)
{
    if(!in_array($request->path(), $this->openRoutes)){
       //middleware code or call of function
    }       

    return $next($request);
}

对于其他中间件，您可以轻松跳过路由文件并根据您的中间件对路由进行分组。

例如：

Route::group(['middleware' => 'checkOnboarding'], function () {
        Route::get('/home', 'HomeController@index');
        Route::get('/account', 'AccountController@index');
    });

Route::group(['prefix' => 'setup'], function () {
    Route::get('/', 'OnboardingController@index')->name('setup');
    Route::post('/settings', 'SettingsController@store');
});

Answer 5

因为 Laravel 7.7 你可以像这样使用 excluded_middleware:

Route::group(['middleware' => ['auth','checkOnboarding']], function () {
    Route::get('/home', 'HomeController@index');
    Route::get('/account', 'AccountController@index');

    Route::group([
      'prefix' => 'setup',
      'excluded_middleware' => ['checkOnboarding'],
], function () {
        Route::get('/', 'OnboardingController@index')->name('setup');
        Route::post('/settings', 'SettingsController@store');
    }); 
});

Answer 6

在Laravel 8.x中你也可以使用withoutMiddleware()方法排除一个或多个路由到组中间件

Route::middleware('auth')->group(function () {
    Route::get('/edit/{id}',[ProgramController::class, 'edit'])->name('edit');

    Route::get('/public', [ProgramController::class, 'public'])
          ->name('public')->withoutMiddleware(['auth']);
});

同时查看官方文档：Here

将中间件应用于除 setup/* 之外的所有路由 Laravel 5.4

Apply Middleware to all routes except `setup/*` in Laravel 5.4

php

laravel

laravel-5

laravel-middleware

laravel-5.4