登录表单问题-Symfony 2.7

我正在遵循Symfony的书和食谱，我遇到了简单登录表单的问题-无论输入的登录/通行证是否有效，都会显示消息-"无效凭据"。用户通过Doctrine（实现UserInterface的用户类）加载。源代码：

安全文件：

providers:
    user_provider:
      entity:
        class: BakaMainBundle:User
firewalls:
    dev:
        pattern: ^/(_(profiler|wdt)|css|images|js)/
        security: false
    default:
        anonymous: ~
        http_basic: ~
        provider: user_provider
        form_login:
           login_path: /login
           check_path: /login_check
           target_path_parameter: /index/welcome
access_control:
  - { path: ^/admin, roles: ROLE_ADMIN }
encoders:
  Baka'MainBundle'Entity'User:
    algorithm: bcrypt
    cost: 12

控制器：

class SecurityController extends Controller
{
    /**
     * @Route("/login", name="login_route")
     */
    public function loginAction()
    {
        $authUtils = $this->get('security.authentication_utils');
        $error = $authUtils->getLastAuthenticationError();
        $enteredUsername = $authUtils->getLastUsername();
        return $this->render('BakaMainBundle::Login.html.twig',
            array
            (
                'last_username' => $enteredUsername,
                'error' => $error,
                'site' => 'login'
            ));
    }
    /**
     * @Route("/login_check", name="login_check")
     */
    public function loginCheckAction()
    {
    }
}

用户存储库：

class UserRepository extends 'Doctrine'ORM'EntityRepository implements UserProviderInterface
{
    public function loadUserByUsername($username)
    {
        $user = $this->createQueryBuilder('u')
            ->where('u.username = :username OR u.email = :email')
            ->setParameter('username', $username)
            ->setParameter('email', $username)
            ->getQuery()
            ->getOneOrNullResult();
        if ($user === null)
        {
            $returnMessage = sprintf(
                '%s - such username of email adress does not exist in database! Try again with other login data.',
                $username);
            throw new UnsupportedUserException($returnMessage);
        }
        return $user;
    }
    public function refreshUser(UserInterface $user)
    {
        $userClass = get_class($user);
        if (!$this->supportsClass($userClass))
        {
            throw new UnsupportedUserException
            (sprintf('Ops! Something goes wrong. Your user class is not supported by security system.'));
        }
        return $this->find($user->getId());
    }
    public function supportsClass($userclass)
    {
        return $this->getEntityName() === $userclass || is_subclass_of($userclass, $this->getEntityName());
    }

和表单html标签：

<form action="{{ path('login_check') }}" method="post">

有什么建议吗？如果能解决我的问题，我将不胜感激。