Php 具有2个参数的拉威尔路线

我有路线结构：

Route::group(array('before' => 'auth'), function() {
    Route::controller('/app/{companyId}', 'AppController');
    Route::controller('/app/{companyId}/projects/{projectId}', 'ProjectsController');
    Route::controller('/app/{companyId}/task/{taskId}', 'TasksController');
    Route::controller('/app/{companyId}/bugs/{bugId}', 'BugsController');
    Route::controller('/app/{companyId}/comments/{commentId}', 'CommentsController');
    Route::controller('/app/{companyId}/calendar', 'CalendarController');
    Route::controller('/', 'CompaniesController');
});

但当我试图打开/app/3/project/1时，它不起作用：

Symfony \ Component \ HttpKernel \ Exception \ NotFoundHttpException
Controller method not found.

什么会导致这个问题

正确的解决方案：检查路线的正确顺序（最长的mut位于顶部）：

Route::controller('app/{companyId}/project/{projectId}', 'ProjectsController');
Route::controller('app/{companyId}/task/{taskId}', 'TasksController');
Route::controller('/app/{companyId}/bugs/{bugId}', 'BugsController');
Route::controller('/app/{companyId}/comments/{commentId}', 'CommentsController');
Route::controller('/app/{companyId}/calendar', 'CalendarController');
Route::controller('/app/{companyId}', 'AppController');
Route::controller('/', 'CompaniesController');

试试这个：

首先，您必须指定路由方法（get或post），然后必须使用@join连接，操作名称将在指定的控制器中执行

Route::get('/app/{companyId}/bugs/{bugId}', 'BugsController@actionname');

---

一点点的实验，我发现了一个错误。原因在于路线的位置。这很有效

Route::controller('app/{companyId}/project/{projectId}', 'ProjectsController');
Route::controller('app/{companyId}/task/{taskId}', 'TasksController');
Route::controller('/app/{companyId}/bugs/{bugId}', 'BugsController');
Route::controller('/app/{companyId}/comments/{commentId}', 'CommentsController');
Route::controller('/app/{companyId}/calendar', 'CalendarController');
Route::controller('/app/{companyId}', 'AppController');
Route::controller('/', 'CompaniesController');

---

谢谢

这不是使用Laravel路由系统的正确方法。当您真正想要命名路由（如检查您的答案）作为解决方案时，您正在尝试这样做，这对其他用户很有用：）