路由问题，基于URL中的变量调用控制器-Laravel4 - Routing Issue, Calling Controller based on Variables in URL- Laravel 4

Routing Issue, Calling Controller based on Variables in URL- Laravel 4

我正在用Laravel 4开发一个应用程序，我需要做的是：假设我有以下路线：

  Route::get('/myroute/{entity}/methodname',
  );

在里面，我需要根据实体变量来决定应该调用哪个控制器和方法，例如：

 'MyNameSpace'MyPackage'StudentController@methodname'

如果

entity == Student

并呼叫

  'MyNameSpace'MyPackage'StaffController@methodname'

如果

    entity == Staff

如何在Laravel 4路由中完成是可能的，还是我必须想出两个不同的路由？

    Route::get('/myroute/Student/methodname') and Route::get('/myroute/Staff/methodname')

这应该符合您的需求

Route::get('/myroute/{entity}/methodname', function($entity){
    $controller = App::make('MyNameSpace''MyPackage'''.$entity.'Controller');
    return $controller->callAction('methodname', array());
}

现在为了避免错误，让我们也检查控制器和操作是否存在：

Route::get('/myroute/{entity}/methodname', function($entity){
    $controllerClass = 'MyNameSpace''MyPackage'''.$entity.'Controller';
    $actionName = 'methodname';
    if(method_exists($controllerClass, $actionName.'Action')){
        $controller = App::make($controllerClass);
        return $controller->callAction($actionName, array());
    }
}

更新

为了使过程更加自动化，你甚至可以将动作名称设置为动态

Route::get('/myroute/{entity}/{action?}', function($entity, $action = 'index'){
    $controllerClass = 'MyNameSpace''MyPackage'''.$entity.'Controller';
    $action = studly_case($action) // optional, converts foo-bar into FooBar for example
    $methodName = 'get'.$action; // this step depends on how your actions are called (get... / ...Action)
    if(method_exists($controllerClass, $methodName)){
        $controller = App::make($controllerClass);
        return $controller->callAction($methodName, array());
    }
}