我编写了自己的密码编码器,它实现了PasswordEncoderInterface
:
class BCryptPasswordEncoder implements PasswordEncoderInterface {
protected $encoder;
public function __construct(BCryptEncoder $encoder) {
$this->encoder = $encoder;
}
public function encodePassword($raw, $salt) {
return $this->encoder->encodeString($raw, $salt);
}
public function isPasswordValid($encoded, $raw, $salt) {
return $this->encoder->encodeString($raw, $salt) == $encoded;
}
}
编码器注册为 id 的服务bcrypt.password.encoder
。但我不知道,如何告诉 symfony 使用它。目前app/config/security.yml
看起来像这样:
security:
encoders:
Symfony\Component\Security\Core\User\User: plaintext
role_hierarchy:
ROLE_ADMIN: ROLE_USER
ROLE_SUPER_ADMIN: [ROLE_USER, ROLE_ADMIN, ROLE_ALLOWED_TO_SWITCH]
providers:
neo4j:
id: security.user.provider.neo4j
firewalls:
dev:
pattern: ^/(_(profiler|wdt)|css|images|js)/
security: false
secured_area:
provider: neo4j
pattern: ^/.*
form_login:
check_path: /login_check
login_path: /login
logout:
path: /logout
target: /
anonymous: ~
access_control:
- { path: ^/login, role: IS_AUTHENTICATED_ANONYMOUSLY }
- { path: ^/.*, role: ROLE_ADMIN }
顺便说一句,我没有使用任何学说实体。
编辑:Symfony\Component\Security\Core\User\User
是我的用户对象。我修改security.yml
了一下:
encoders:
Symfony\Component\Security\Core\User\User:
id: bcrypt.password.encoder
这会导致致命错误:
Catchable Fatal Error: Argument 1 passed to EMC3\Bundle\UserBundle\Neo4jUserProvider::__construct() must be an instance of EMC3\Bundle\UserBundle\UserManager, instance of EMC3\Bundle\UserBundle\BCryptEncoder given, called in /var/www/emc3/app/cache/dev/appDevDebugProjectContainer.php on line 227 and defined in /var/www/emc3/src/EMC3/Bundle/UserBundle/Neo4jUserProvider.php line 29
这对我来说没有任何意义。