Symfony UserVoter在url中获取当前登录用户而不是User

时间:2017-06-19 16:33:51

标签: symfony symfony-3.2 symfony2-voter

在show user操作中,我想检查记录的用户是否有权查看此用户。所以我创建了一个UserVoter。 但是当我尝试使用注释将url中定义的用户传递给Voter时,我使用$subject$token->getUser()

来获取记录的用户

如果我在动作中更改我的var名称,它可以正常工作($ user - > $ foo)。

您知道如何更改我的var名称吗?

控制器:

/**
 * Finds and displays a user entity.
 *
 * @Method("GET")
 * @Route("/{id}/", name="authenticated_user_show", requirements={"id": "\d+"})
 * @ParamConverter("user", class="AppBundle:User")
 * @Security("is_granted('SHOW', user)")
 */
public function showAction(User $user)
{

选民:

protected function voteOnAttribute($attribute, $subject, TokenInterface $token)
{
    $user = $token->getUser();

    if (!$user instanceof User) {
        // the user must be logged in; if not, deny access
        return false;
    }

    if ($user->hasRole(User::ROLE_SUPER_ADMIN)) {
        // if super admin, can do everything
        return true;
    }

    // you know $subject is an User object, thanks to supports
    /** @var User $userSubject */
    $userSubject = $subject;

    switch ($attribute) {
        case self::SHOW:
            return $this->canShow($userSubject, $user);
        case self::EDIT:
            return $this->canEdit($userSubject, $user);
    }

    throw new \LogicException('This code should not be reached!');
}


private function canShow(User $userSubject, User $user) : bool
{
    if ($user->getClient() === $userSubject->getClient()) {
        // if they are in the same client
        return true;
    }

    return false;
}

private function canEdit(User $userSubject, User $user, TokenInterface $token) : bool
{
    if (
        $this->decisionManager->decide($token, [User::ROLE_ADMIN]) &&
        $user->getClient() === $userSubject->getClient()
        ) {
        // if the user and the admin belong to the same client
        return true;
    } elseif (
        $this->decisionManager->decide($token, [User::ROLE_MANAGER]) &&
        $this->userManager->hasEstablishmentInCommon($user, $userSubject)
        ) {
        // if the user and the manager are linked to the same establishment
        return true;
    }

    return false;
}

1 个答案:

答案 0 :(得分:0)

假设您没有尝试查看已记录的用户,我会认为您可能与安全注释上下文中的变量存在冲突:

https://symfony.com/doc/current/bundles/SensioFrameworkExtraBundle/annotations/security.html 安全注释可以访问以下变量:

  • 令牌:当前安全令牌
  • user:当前用户对象
  • request:请求实例
  • 角色:用户角色;和所有要求 属性。

要修复此问题,请更改用户名,以免发生冲突。

/**
 * Finds and displays a user entity.
 *
 * @Method("GET")
 * @Route("/{id}/", name="authenticated_user_show", requirements={"id": 
   "\d+"})
 * @ParamConverter("showUser", class="AppBundle:User")
 * @Security("is_granted('SHOW', showUser)")
 */
 public function showAction(User $showUser)