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Symfony2使用api密钥访问控制问题对用户进行身份验证
EN

Stack Overflow用户
提问于 2015-09-11 11:55:26
回答 1查看 405关注 0票数 1

嗨,我正在为自己开发一个项目。我想要做的是验证除用户注册之外的所有api调用。我创建了用户身份验证和用户提供程序,如以下链接所述:http://symfony.com/doc/current/cookbook/security/api_key_authentication.html

一切正常,所有到我后端的urls都经过了身份验证,它们由UserTokenAuthenticator和UserTokenProvider处理。但它验证每个呼叫的东西。我不想验证此url

代码语言:javascript
复制
    /v1.0/users with POST method

因此,我在security.yml文件中设置了如下访问控制:

代码语言:javascript
复制
access_control:
    user_register:
        path: ^/v1.0/users
        roles: IS_AUTHENTICATED_ANONYMOUSLY
        methods: [POST]

但是每当我尝试对这个url进行post调用时,它都会尝试进行身份验证(bec调用没有将token作为参数),但它失败了。如何防止仅对此路由进行身份验证??下面是我的全部代码:

security.yml:

代码语言:javascript
复制
security:
    encoders:
        entity_user:
            class: KBell\AppBundle\Entity\User
            algorithm: sha1
            iterations: 1
            encode_as_base64: false
        entity_device:
            class: KBell\AppBundle\Entity\Device
            algorithm: sha1
            iterations: 1
            encode_as_base64: false

    role_hierarchy:
        ROLE_DEVICE: ROLE_DEVICE
        ROLE_USER : ROLE_USER

    providers:
        entity_device:
            entity:
                class: KBell\AppBundle\Entity\Device
                property: name
        entity_user:
            entity:
                class: KBell\AppBundle\Entity\User
                property: email
        user_token_provider:
            id: user_token_provider
    firewalls:
        dev:
            pattern: ^/(_(profiler|wdt)|css|images|js)/
            security: false
        api_user_secured_area:
            pattern: ^/v1.0/users
            stateless: true
            simple_preauth:
                authenticator: user_token_authenticator
            provider: user_token_provider
    access_control:
        user_register:
            path: ^/v1.0/users
            roles: IS_AUTHENTICATED_ANONYMOUSLY
            methods: [POST]

下面是UserTokenAuthenticator:

代码语言:javascript
复制
<?php

class UserTokenAuthenticator implements SimplePreAuthenticatorInterface,    AuthenticationFailureHandlerInterface
{

    public function createToken(Request $request, $providerKey)
    {
        $accessToken = $request->query->get('access_token');
        if (!$accessToken) {
            throw new BadCredentialsException('Access Token is missing');
        }

        return new PreAuthenticatedToken(
            'anon.',
            $accessToken,
            $providerKey
        );
    }

    public function authenticateToken(TokenInterface $token, UserProviderInterface $userProvider, $providerKey)
    {
        $accessToken = $token->getCredentials();
        $user = $userProvider->getUserForAccessToken($accessToken);
        if (!$user) {
            throw new AuthenticationException(
                sprintf('Access Token "%s" does not exist.', $accessToken)
            );
        }

        return new PreAuthenticatedToken(
            $user,
            $accessToken,
            $providerKey,
            $user->getRoles()
        );
    }

    public function supportsToken(TokenInterface $token, $providerKey)
    {
        return $token instanceof PreAuthenticatedToken && $token->getProviderKey() === $providerKey;
    }

    public function onAuthenticationFailure(Request $request, AuthenticationException $exception)
    {
        return ApiUtil::getJsonErrorResponse("User authentication is failed: ".$exception->getMessage(), Response::HTTP_UNAUTHORIZED, "AuthenticationException");
    }
}

下面是UserTokenProvider.php:

代码语言:javascript
复制
class UserTokenProvider implements UserProviderInterface
{
    private $userManager;

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

    public function getUserForAccessToken($accessToken)
    {
        $user = $this->userManager->findUserByAccessToken($accessToken);
        return $user;
    }

    public function loadUserByUsername($username)
    {
    }

    public function refreshUser(UserInterface $user)
    {
        throw new UnsupportedUserException();
    }

    public function supportsClass($class)
    {
        return 'Symfony\Component\Security\Core\User\User' === $class;
    }
}

我为此花了很多时间。任何帮助都将不胜感激!谢谢!

EN

回答 1

Stack Overflow用户

发布于 2015-09-11 14:25:59

我认为你可以尝试这样配置你的防火墙设置:

代码语言:javascript
复制
firewalls:
    secured_user_area:
        pattern: /v(\d+\.?\d*)/(?!users)
        stateless: true
        provider: user_token_provider

    ....

在这里,?!users (或者甚至像?!users|URL2|URL3这样的多个urls )声明将urls排除在防火墙之外。

票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/32515191

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