自定义身份验证提供程序中的 Symfony2 ContextErrorException

Question

我正在尝试创建一个How to create a custom Authentication Provider，在完整阅读并查看了 Symfony 代码之后，我认为只需创建一个工厂、身份验证提供程序并使用 symfony 默认类就足够了，但实际上我是遗漏了一些东西，我收到了这个错误

ContextErrorException: Catchable Fatal Error: Argument 1 passed to Acme\DemoBundle\Provider\MyProvider::__construct() must implement interface Symfony\Component\Security\Core\User\UserProviderInterface, string given, called in D:\wamp\www\sf2ldap\app\cache\dev\appDevDebugProjectContainer.php on line 1383 and defined in D:\wamp\www\sf2ldap\src\Acme\DemoBundle\Provider\MyProvider.php line 20

工厂

namespace Acme\DemoBundle\Factory;

use Symfony\Component\DependencyInjection\ContainerBuilder;
use Symfony\Component\DependencyInjection\Reference;
use Symfony\Component\DependencyInjection\DefinitionDecorator;
use Symfony\Component\Config\Definition\Builder\NodeDefinition;
use Symfony\Bundle\SecurityBundle\DependencyInjection\Security\Factory\AbstractFactory;

class MyFactory extends AbstractFactory
{

    public function getPosition()
    {
        return 'form';
    }

    public function getKey()
    {
        return 'kstr';
    }

    protected function createAuthProvider(ContainerBuilder $container, $id, $config, $userProviderId)
    {

        $providerId = 'security.authentication.provider.kstr.' . $id;
        $container
                ->setDefinition($providerId, new DefinitionDecorator('kstr.security.authentication.provider'))
                ->replaceArgument(0, new Reference($userProviderId));
        return $providerId;
    }

    protected function getListenerId()
    {
        return 'security.authentication.listener.form';
    }

}

我的提供者

namespace Acme\DemoBundle\Provider;

use Symfony\Component\Security\Core\Authentication\Provider\AuthenticationProviderInterface;
use Symfony\Component\Security\Core\User\UserProviderInterface;
use Symfony\Component\Security\Core\Exception\AuthenticationException;
use Symfony\Component\Security\Core\Authentication\Token\TokenInterface;

class MyProvider implements AuthenticationProviderInterface
{

    private $_userProvider;

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

    public function authenticate(TokenInterface $token)
    {
        try
        {
            $user = $this->_userProvider->loadUserByUsername($token->getUsername());
            //custom auth steps
            $token = new UsernamePasswordToken(
                    $token->getUsername(), null, $token->getProviderKey(), $user->getRoles()
            );
            return $token;
            }
        } catch (\Exception $exc)
        {
            throw new AuthenticationException('Invalid username or password. ', 0, $e);
        }
        throw new AuthenticationException('Invalid username or password asdfasd');
    }

    public function supports(TokenInterface $token)
    {
        return $token instanceof UsernamePasswordToken;
    }

}

services.yml

services:
    kstr.security.authentication.provider:
        class:  Acme\DemoBundle\Provider\MyProvider
        arguments: [""]

security.yml

security:
    encoders:
        Acme\DemoBundle\Entity\SecureUser: plaintext
    providers:
        multiples:
            chain:
                providers: [entity_provider, ldap]
        entity_provider:
          entity: { class: AcmeDemoBundle:SecureUser, property: username }
        ldap:
          id: kstr.security.authentication.provider

    firewalls:
        dev:
            pattern:  ^/(_(profiler|wdt)|css|images|js)/
            security: false

        login:
            pattern:  ^/demo/secured/login$
            security: false

        secured_area:
            pattern:    ^/demo/secured/
            kstr: 
                check_path: _security_check
                login_path: _demo_login
                provider: ldap
            logout:
                path:   _demo_logout
                target: _demo

需要一些帮助来解决这个问题，我在这里缺少什么？即使默认的“security.authentication.listener.form”满足我的需求，我是否需要创建自定义侦听器？

Answer 1

您将 字符串 arguments: [""] 作为第一个参数传递给服务的构造函数。

这就是__construct(UserProviderInterface $userProvider) 中的类型提示失败的原因。

正确注入UserProviderInterface，异常就会消失。