我有一个用symfony2开发的服务器端api,现在我尝试进行身份验证。
客户端/移动设备应用程序必须使用API密钥进行身份验证
使用应用程序的用户必须使用电子邮件+密码进行身份验证并获得访问令牌
因此我使用这个防火墙和apikey验证器

firewalls:
    login:
        pattern:  ^/login$
        security: false

    secured_area:
        pattern: ^/
        stateless: true
        simple_preauth:
            authenticator: apikey_authenticator

API密钥验证器
namespace Rental\APIBundle\Security;

use Symfony\Component\Security\Core\Authentication\SimplePreAuthenticatorInterface;
use Symfony\Component\Security\Core\Authentication\Token\TokenInterface;
use Symfony\Component\Security\Core\Exception\AuthenticationException;
use Symfony\Component\Security\Core\Authentication\Token\PreAuthenticatedToken;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\Security\Core\User\UserProviderInterface;
use Symfony\Component\Security\Core\Exception\UsernameNotFoundException;
use Symfony\Component\Security\Core\Exception\BadCredentialsException;


class ApiKeyAuthenticator implements SimplePreAuthenticatorInterface    {
    protected $userProvider;

    public function __construct(ApiKeyUserProvider $userProvider)
    {
        $this->userProvider = $userProvider;
    }

    public function createToken(Request $request, $providerKey)
    {

        //$apiKey = $request->query->get('apikey');
        // use test value
        $apiKey = "234234234";

        if (!$apiKey) {
            throw new BadCredentialsException('No API key found');
        }

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

    public function authenticateToken(TokenInterface $token, UserProviderInterface $userProvider, $providerKey)
    {
        $apiKey = $token->getCredentials();
        $username = $this->userProvider->getUsernameForApiKey($apiKey);

        if (!$username) {
            throw new AuthenticationException(
                sprintf('API Key "%s" does not exist.', $apiKey)
            );
        }

        $user = $this->userProvider->loadUserByUsername($username);

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

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

到目前为止,没问题。现在这个类使用以下类的方法
namespace Rental\APIBundle\Security;

use Symfony\Component\Security\Core\User\UserProviderInterface;
use Symfony\Component\Security\Core\User\User;
use Symfony\Component\Security\Core\User\UserInterface;
use Symfony\Component\Security\Core\Exception\UnsupportedUserException;

class ApiKeyUserProvider implements UserProviderInterface {
    public function getUsernameForApiKey($apiKey)
    {
        // Look up the username based on the token in the database
        // use test value
        $username = "Alex";

        return $username;
    }

    public function loadUserByUsername($username)
    {
        // return User by Username
    }

    public function refreshUser(UserInterface $user)
    {
        // code
    }

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

我的具体问题是:
方法需要通过搜索用户名来查找用户实体。但从这个类我没有访问数据库的权限。我发现了使用静态方法loadUserByUsername的示例,但是没有这种方法,而且实体(mvc的模型)也没有访问数据库的权限。如何将用户从数据库中取出?
我想首先认证应用程序本身,以便进行一般的api访问,其次认证用户,以获取个人信息和有限的编辑权限。当用户通过电子邮件和密码登录时,数据被保存,例如在usernamepassworttoken中,来自同一客户端的下一个调用如何使用访问令牌访问数据。会话对这些ajax http请求没有影响。

最佳答案

一。
您应该对它使用依赖注入,并在您的提供者中注入实体管理器。

 your_api_key_user_provider:
            class:     Rental\APIBundle\Security\ApiKeyUserProvider
            arguments: ["@doctrine.orm.entity_manager"]
 apikey_authenticator:
            class:     Rental\APIBundle\Security\ApiKeyAuthenticator
            arguments: [""@your_api_key_user_provider"]

之后,将其添加到提供程序中:
    use Doctrine\ORM\EntityManager;

    class ApiKeyUserProvider implements UserProviderInterface {

        protected $em;

        public function __construct(EntityManager $em){
            $this->em = $entityManager;
        }
       //... Now you have access to database

    }

2.ajax可以发送cookies,php可以像处理普通会话和use会话一样处理这些请求。确保您的请求是发送cookies(请参见Why is jquery's .ajax() method not sending my session cookie?

09-11 19:37
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