symfony2 : check isGranted for a route

Question

假设有某些路由字符串，例如受防火墙保护的“/path/index.html”，如何检查当前用户是否能够访问它？

提前致谢!

对不起，我应该更明确一点:我有一个路由名称数组，我构建了一个菜单。许多具有不同角色的用户可以使用此菜单访问页面。目的是仅在此菜单中为特定用户显示可访问的喜欢。

就像是:

'security_context'->'user'->isGranted('/path/index.html')

Answer 1

此答案基于您的评论:
您应该获得访问该路由所需的角色。您需要访问 security.access_map服务是私有(private)的。所以它必须直接注入(inject)。例如:你可以创建一个 path_roles像这样的服务，您可以获得特定路径的角色:

namespace Acme\FooBundle;

class PathRoles
{
    protected $accessMap;

    public function __construct($accessMap)
    {
        $this->accessMap = $accessMap;
    }

    public function getRoles($path)
    { //$path is the path you want to check access to

        //build a request based on path to check access
        $request = Symfony\Component\HttpFoundation\Request::create($path, 'GET');
        list($roles, $channel) = $this->accessMap->getPatterns($request);//get access_control for this request

        return $roles;
    }
}

现在将其声明为服务:

services:
    path_roles:
        class: 'Acme\FooBundle\PathRoles'
        arguments: ['@security.access_map']

现在在您的 Controller 中使用该服务来获取路径的角色并根据这些角色和 isGranted.i.e 构建您的菜单:

  //code from controller
  public function showAction(){
      //do stuff and get the link path for the menu,store it in $paths
      $finalPaths=array();
      foreach($paths as $path){
      $roles = $this->get('path_roles')->getRoles($path);
      foreach($roles as $role){
          $role = $role->getRole();//not sure if this is needed
          if($this->get('security.context')->isGranted($role)){
              $finalPaths[] = $path;
              break;
          }
      }
     //now construct your menu based on $finalPaths
      }
  }