Laravel 5 路由组中过滤器的异常路由

Question

我有一个简单的问题，假设我有一个如下所示的溃败组：

/*
|---------------------------------------------------------------------------
|   The following correspondent routes are namespaced to 'Profile'
|   and all controlles under the respective folder.
|
|   Routes are protected by milldeware 'auth'
|---------------------------------------------------------------------------
*/
Route::group(['middleware' => 'auth', 'namespace' => 'Profile'], function(){
/* Do something here */ 
});

如您所见，路由受中间件 'auth' 保护，我想问的是，如果我不希望它受中间件保护的那组路由很少怎么办，我应该如何陈述例外规则？我可以这样做吗？

/*
|---------------------------------------------------------------------------
|   The following correspondent routes are namespaced to 'Profile'
|   and all controlles under the respective folder.
|
|   Routes are protected by milldeware 'auth'
|---------------------------------------------------------------------------
*/
Route::group(['middleware' => 'auth', 'namespace' => 'Profile'], function(){
/* Do something here */ 
});


/*
|---------------------------------------------------------------------------
|   The following correspondent routes are namespaced to 'Profile'
|   and all controlles under the respective folder.
|
|   Routes are not protected by milldeware 'auth'
|---------------------------------------------------------------------------
*/
Route::group('namespace' => 'Profile', function(){
/* Do something unprotected here */ 
});

或者有合适的方法吗？

Answer 1

正确的方法是这样嵌套路由组：

Route::group(['namespace' => 'Profile'], function() {

    // Your non protected routes go here
    // ...

    Route::group(['middleware' => 'auth'], function(){
        // Your protected routes go here
        // ...
    });
});

嵌套路由组时，每个子组都会继承父组的所有参数。

Laravel 5 路由组中过滤器的异常路由

Laravel 5 exception routes from filter in route groups

laravel

laravel-routing

laravel-5