Question

我在symfony 2.8项目中。另一位开发人员编写了一个Voter来检查用户是否具有特定任务的权限。我已经在控制器中使用此功能而没有任何问题，但现在我正在编写服务以获取动态菜单而我不知道如何访问方法isGranted：

错误：尝试调用名为＆＃34; isGranted＆＃34;的未定义方法。的类。

namespace NameSpaceQuestion\Question\Service;

use Doctrine\ORM\EntityManager;

class MenuBuilder
{

    private $areasTools;
    public $menuItems = array();
    private $em;

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

    public function createMenuCourse($course,$mode,$user)
    {
        $repository = $this->em->getRepository('eBundle:AreasTools');
        $areas = $repository->findAll();
        //Retrieve every area from the db
        $itemsMenu = array();

        foreach ($areas as $area) {

            //If the user has permissions the area is included in the menu and proceed to check the tools that belong to the current area
            if($this->isGranted( $mode,$area,$user ) ){
                $itemsMenu[$area->getName()] = array();
                $toolsPerCourse = $this->em->getRepository('eBundle:CourseTool')->findByAreaToolAndCourse($area, $course);
                foreach ($toolsPerCourse as $toolCourse) {
                    //If the user has permissions the tool is included in the menu under the respective Area
                    if( ($this->isGranted( $mode,$toolCourse,$user ) )){
                        array_push($itemsMenu[$area->getName()], $toolCourse->getName());
                    }
                }

            }
        }

        return $itemsMenu;
    }
}

Answer 1

您必须使用Dependency Injection在MenuBuilder类中获取AuthorizationChecker。您可以在此处阅读：http://symfony.com/doc/current/cookbook/security/securing_services.html

因为看起来你已经注入了EntityManager，只需将AuthorizationChecker添加到你的代码中：

use Symfony\Component\Security\Core\Authorization\AuthorizationCheckerInterface;

class MenuBuilder
{

    protected $authorizationChecker;

    public function __construct(EntityManager $em, AuthorizationCheckerInterface $authorizationChecker)
    {
        $this->authorizationChecker = $authorizationChecker;
        $this->em = $em;
    }

    function createMenuCourse()
    {
         if ( $this->authorizationChecker->isGranted('EDIT',$user) ) {
               //build menu
         }
    }
}

Answer 2

首先要了解的是Symfony基本控制器有许多辅助函数，例如isGranted实现。但是，当然，如果您不在控制器内，那么您无法访问它们。另一方面，查看基类并复制出所需的功能是有益的。

isGranted功能依赖于授权检查服务，您需要将其注入菜单构建器，从而产生如下内容：

class MenuBuilder
{
    private $em;
    private $authorizationChecker;

    public function __construct(
        EntityManager $em, 
        $authorizationChecker // security.authorization_checker
    ) {
        $this->em = $em;
        $this->authorizationChecker = $authorizationChecker;
    }    
    protected function isGranted($attributes, $object = null)
    {
        return $this->authorizationChecker->isGranted($attributes, $object);
    }

认真。斯蒂芬在一分钟之内打败了我。必须学会更快地输入。

如何在服务中访问选民方法

2 个答案: