如果用户状态为非活动状态,则阻止用户登录
Posted
技术标签:
【中文标题】如果用户状态为非活动状态,则阻止用户登录【英文标题】:Prevent user from logging in if his status is inactive 【发布时间】:2020-06-06 00:08:46 【问题描述】:我试图阻止用户登录他的状态是非活动的。我正在使用带有 LexikJWT 捆绑包的 API 平台。
我试图通过扩展JWTTokenAuthenticator->checkCredentials
来创建JWTAuthentication
保护,但问题是这在用户已经登录后才有效。
我想要实现的是向用户返回一条消息,他需要首先激活他的帐户,或者任何其他消息,最好是在任何自定义条件下的任何自定义消息。
我的安全 YAML 如下所示:
security:
encoders:
App\Entity\User:
algorithm: bcrypt
providers:
app_user_provider:
entity:
class: App\Entity\User
property: email
firewalls:
dev:
pattern: ^/_(profiler|wdt)
security: false
api:
pattern: ^/api/
stateless: true
anonymous: true
provider: app_user_provider
json_login:
check_path: /api/authentication_token
username_path: email
password_path: password
success_handler: lexik_jwt_authentication.handler.authentication_success
failure_handler: lexik_jwt_authentication.handler.authentication_failure
guard:
authenticators:
- app.jwt_token_authenticator
main:
anonymous: true
access_control:
- path: ^/api/authentication_token, roles: IS_AUTHENTICATED_ANONYMOUSLY
- path: ^/api/graphql, roles: IS_AUTHENTICATED_ANONYMOUSLY
- path: ^/public-api, roles: IS_AUTHENTICATED_ANONYMOUSLY
- path: ^/api/, roles: [ROLE_MANAGER, ROLE_LEADER]
- path: ^/, roles: IS_AUTHENTICATED_ANONYMOUSLY
服务:
services:
# default configuration for services in *this* file
_defaults:
autowire: true # Automatically injects dependencies in your services.
autoconfigure: true # Automatically registers your services as commands, event subscribers, etc.
# makes classes in src/ available to be used as services
# this creates a service per class whose id is the fully-qualified class name
App\:
resource: '../src/*'
exclude: '../src/DependencyInjection,Entity,Migrations,Tests,Kernel.php'
# controllers are imported separately to make sure services can be injected
# as action arguments even if you don't extend any base controller class
App\Controller\:
resource: '../src/Controller'
tags: ['controller.service_arguments']
# add more service definitions when explicit configuration is needed
# please note that last definitions always *replace* previous ones
gedmo.listener.softdeleteable:
class: Gedmo\SoftDeleteable\SoftDeleteableListener
tags:
- name: doctrine.event_subscriber, connection: default
calls:
- [ setAnnotationReader, [ '@annotation_reader' ] ]
acme_api.event.authentication_success_listener:
class: App\EventListener\AuthenticationSuccessListener
tags:
- name: kernel.event_listener, event: lexik_jwt_authentication.on_authentication_success, method: onAuthenticationSuccessResponse
app.jwt_token_authenticator:
autowire: false
autoconfigure: false
class: App\Security\Guard\JWTTokenAuthenticator
parent: lexik_jwt_authentication.security.guard.jwt_token_authenticator
'App\Serializer\ApiNormalizer':
decorates: 'api_platform.serializer.normalizer.item'
arguments: ['@App\Serializer\ApiNormalizer.inner', '@doctrine.orm.entity_manager']
'App\Serializer\HydraApiNormalizer':
decorates: 'api_platform.jsonld.normalizer.item'
arguments: ['@App\Serializer\ApiNormalizer.inner', '@doctrine.orm.entity_manager']
'App\Voter\ModifyUserVoter':
public: false
tags:
- name: security.voter
身份验证器保护
class JWTTokenAuthenticator extends BaseAuthenticator
/**
* @inheritdoc
*/
public function checkCredentials($credentials, UserInterface $user)
if (!$user->getRoles() || !in_array($user->getRoles()[0], ['ROLE_MANAGER', 'ROLE_LEADER']))
throw new UnauthorizedHttpException(rand(10000, 99999), 'Unauthorized');
if (!$user->getStatus() != "active")
throw new UnauthorizedHttpException(rand(10000, 99999), 'Unauthorized');
return true;
【问题讨论】:
【参考方案1】:通过在User
实体上实现AdvancedUserInterface
而不是UserInterface
并向isEnabled()
方法添加逻辑,我设法实现了我想要的。
【讨论】:
AdvancedUserInterface
自 4.1 起已弃用,并在 5 上删除。检查其他答案以向前兼容的方式进行。【参考方案2】:
您需要创建UserCheckerInterface
的实现。 (Docs)
例如,看这个:
use Symfony\Component\Security\Core\Exception\DisabledException;
use Symfony\Component\Security\Core\User\UserCheckerInterface;
use Symfony\Component\Security\Core\User\UserInterface;
class EasyUserChecker implements UserCheckerInterface
public function checkPreAuth(UserInterface $user): void
// my checker only cares for our managed user classes, we return with no action
if (!$user instanceof AppAdmin && !$user instanceof AppUser)
return;
// our user entities can be deleted or disabled. If the user is neither, we return with no action
if (!$user->isDeleted() && !empty($user->isEnabled()))
return;
// if we got here, we throw an exception
throw new DisabledException('User account is disabled.');
// I'm not using the post authorization check, but needs to have an implementation to satisfy the interface.
public function checkPostAuth(UserInterface $user): void
您在安全配置中启用检查器。例如:
security:
firewalls:
api:
pattern: ^/api
user_checker: App\Security\EasyChecker
您现在不应该编写 AdvancedUserInterface
的新实现。将其用作解决方案是错误的方法。
该接口is deprecated since 4.1,并在 Symfony 5 中完全删除。因此依赖于该接口的代码将无法升级到较新的 Symfony 版本。
【讨论】:
谢谢,我会在工作中尝试一下,我会接受你的回答。也谢谢你的样品和解释我真的很感激 谢谢,这按预期工作。我已将 AdvancedUserInterface 替换为 UserInterace。一件小事,登录后我更改了用户的状态,以查看用户登录时的行为,突然他被禁用,然后它可以工作,但它不会像它一样返回传递给 DisabledException 的错误消息登录时,它会返回一些默认消息,我猜有些东西会覆盖它,知道什么或在哪里吗?以上是关于如果用户状态为非活动状态,则阻止用户登录的主要内容,如果未能解决你的问题,请参考以下文章
Laravel 5.4 检查用户是不是处于活动状态,如果没有则回复消息