Laravel中间件除了Route::group

Question

我正在尝试为管理部分创建一个组路由，并将中间件应用于除登录和注销以外的所有路径。

到目前为止，我有：

Route::group(['prefix' => 'admin', 'namespace' => 'Admin', 'middleware' => 'authAdmin'], function() {

    Route::resource('page', 'PageController');
    Route::resource('article', 'ArticleController');
    Route::resource('gallery', 'GalleryController');
    Route::resource('user', 'UserController');

    // ...

});

如何在上述设置中声明中间件的异常?

- Sebastian Sulinski

Answer 1

只需嵌套组，然后您就可以排除特定路由：

Route::group(['prefix' => 'admin', 'namespace' => 'Admin'], function() {

    Route::get('login', 'AuthController@login');
    Route::get('logout', 'AuthController@logout');

    Route::group(['middleware' => 'authAdmin'], function(){
        Route::resource('page', 'PageController');
        Route::resource('article', 'ArticleController');
        Route::resource('gallery', 'GalleryController');
        Route::resource('user', 'UserController');

        // ...
    });
});

- lukasgeiter

Answer 2

您可以使用 Laravel 的 withoutMiddleware 方法，示例代码如下：

Route::group(['prefix' => 'admin', 'namespace' => 'Admin', 'middleware' => 'authAdmin'], function() {

  Route::resource('page', 'PageController');
  Route::resource('article', 'ArticleController');
  Route::resource('gallery', 'GalleryController');
  Route::resource('user', 'UserController');

  Route::get('login', 'AuthController@login')->withoutMiddleware([AuthAdminMiddleware::class]);
  Route::get('logout', 'AuthController@logout')->withoutMiddleware([AuthAdminMiddleware::class]);
});

- Tuncay Elvanağaç