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我按照本指南实现了我自己的自定义用户登录。不幸的是,它Bad credentials在登录时说。这个例外来自第 72 行Symfony\Component\Security\Core\Authentication\Provider\UserAuthenticationProvider。抛出此异常是因为它无法检索用户。

我为自定义需求所做的更改是用户没有用户名。他们将使用他们的电子邮件地址登录。但我认为实施起来没有问题。

安全性.yml:

security:
    encoders:
        Acme\UserBundle\Entity\User: plaintext

    role_hierarchy:
        ROLE_ADMIN:       ROLE_USER
        ROLE_SUPER_ADMIN: [ROLE_USER, ROLE_ADMIN, ROLE_ALLOWED_TO_SWITCH]

    providers:
        administrators:
            entity: { class: AcmeUserBundle:User }

    firewalls:
        secured_area:
            pattern: ^/
            anonymous: ~
            form_login: ~

    access_control:
        - { path: ^/login, roles: IS_AUTHENTICATED_ANONYMOUSLY }
        - { path: ^/register, roles: IS_AUTHENTICATED_ANONYMOUSLY }
        - { path: ^/, roles: ROLE_USER }

用户存储库:

class UserRepository extends EntityRepository implements UserProviderInterface
{
    public function loadUserByUsername($username)
    {
        $q = $this
            ->createQueryBuilder('u')
            ->where('u.email = :email')
            ->setParameter('email', $username)
            ->getQuery();

        try {
            // The Query::getSingleResult() method throws an exception
            // if there is no record matching the criteria.
            $user = $q->getSingleResult();
        } catch (NoResultException $e) {
            $message = sprintf(
                'Unable to find an active admin AcmeUserBundle:User object identified by "%s".',
                $username
            );
            throw new UsernameNotFoundException($message, 0, $e);
        }

        return $user;
    }

    public function refreshUser(UserInterface $user)
    {
        $class = get_class($user);
        if (!$this->supportsClass($class)) {
            throw new UnsupportedUserException(
                sprintf(
                    'Instances of "%s" are not supported.',
                    $class
                )
            );
        }

        return $this->find($user->getId());
    }

    public function supportsClass($class)
    {
        return $this->getEntityName() === $class
        || is_subclass_of($class, $this->getEntityName());
    }
}

login.twig.html:

{% if error %}
    <div>{{ error.message }}</div>
{% endif %}

<form action="{{ path('login_check') }}" method="post">
    <legend>Login</legend>
    <label for="email">Email:</label>
    <input type="email" id="email" name="_email" value="{{ email }}"

    <label for="password">Password:</label>
    <input type="password" id="password" name="_password" />

    <button type="submit">Login</button>
</form>

我在这里做错了什么?其中UserRepository明确将电子邮件作为用户名查询,为什么找不到用户?我猜测它与csrf_token? 如何将其添加到控制器和树枝文件中?这完全是问题吗,还是我做错了什么?

4

1 回答 1

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默认情况下,Symfony 安全使用_username_password参数通过表单提交来验证用户。您可以在安全配置参考中看到它

form_login:
    username_parameter: _username
    password_parameter: _password

所以你需要放置_username字段名称而不是_email.

于 2013-09-04T17:25:19.187 回答