如果已经在其他应用程序中登录主域,如何使用 symfony 安全登录到子域?
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【中文标题】如果已经在其他应用程序中登录主域,如何使用 symfony 安全登录到子域?【英文标题】:how to login to subdomain using symfony security if already loggedin on main domain in other app? 【发布时间】:2017-01-28 07:14:38 【问题描述】:我已经登录到主域。说 example.com(在旧版 kohana 中开发的应用程序)。我正在尝试登录 subdmain,比如说 foo.bar.example.com 。
foo.example.com 是 symfony 应用程序。下面是我的配置。开发工具栏显示“匿名”用户。它不会从 cookie 中的会话 ID 登录用户。
security.yml
# To get started with security, check out the documentation:
# http://symfony.com/doc/current/book/security.html
security:
# http://symfony.com/doc/current/book/security.html#where-do-users-come-from-user-providers
providers:
in_memory:
memory: ~
firewalls:
# disables authentication for assets and the profiler, adapt it according to your needs
dev:
pattern: ^/(_(profiler|wdt)|css|images|js)/
security: false
main:
anonymous: ~
# activate different ways to authenticate
# http_basic: ~
# http://symfony.com/doc/current/book/security.html#a-configuring-how-your-users-will-authenticate
# form_login: ~
# http://symfony.com/doc/current/cookbook/security/form_login_setup.html
Config.yml
framework:
#esi: ~
#translator: fallbacks: ["%locale%"]
secret: "%secret%"
router:
resource: "%kernel.root_dir%/config/routing.yml"
strict_requirements: ~
form: ~
csrf_protection: ~
validation: enable_annotations: true
#serializer: enable_annotations: true
default_locale: "%locale%"
trusted_hosts: ~
trusted_proxies: %trusted_proxies%
session:
# handler_id set to null will use default session handler from php.ini
handler_id: 'snc_redis.session.handler'
name: 'MY_COOKIE_NAME'
cookie_domain: '.example.com'
fragments: ~
http_method_override: true
request:
formats:
pdf: 'application/pdf'
epub: 'application/epub+zip'
snc_redis:
session:
client: session
prefix: ''
clients:
default:
type: predis
alias: default
dsn: %redis_dsn%
logging: false
options:
# profile: 2.6
profile: 2.2
connection_persistent: false
slave:
type: predis
alias: slave
logging: false
dsn: %redis_dsn_slave%
session:
type: predis
alias: session
dsn: %redis_dsn%
我是否需要配置至少一个身份验证提供程序?要么 我需要编写自定义身份验证提供程序,就像记住我一样工作?
Symfony\Component\Security\Http\Firewall\AbstractAuthenticationListener->句柄
有
if ($this->options['require_previous_session'] && !$request->hasPreviousSession())
throw new SessionUnavailableException('Your session has timed out, or you have disabled cookies.');
请求有
public function hasPreviousSession()
// the check for $this->session avoids malicious users trying to fake a session cookie with proper name
return $this->hasSession() && $this->cookies->has($this->session->getName());
【问题讨论】:
默认情况下,会话 cookie 配置在子域上,为了使相同的 cookie 可访问,请确保检查 SESSIONID cookie 是否具有正确的 TLD 域。所以,“example.com”而不是“sub.example.com”。 我已正确设置域。但仍然没有运气 Symfony 不会自动使用SESSIONID
,您应该创建自己的Provider
,以便将该会话与(共享)数据库中的会话相匹配。之后您可以手动设置UsernamePasswordToken
。
必须有默认实现,将 cookie 中的 sessionId 与存储在文件系统中的 session 相匹配。我已经配置了 redis 会话处理程序,并将其映射到 session.handler_id。抱歉无法理解为什么我们需要编写自定义提供程序。
【参考方案1】:
在主域和登录后,您可以将令牌保存到数据库和 cookie 中,在第二个应用程序中,您应该创建一个自定义身份验证器,该身份验证器可以从导航器中恢复令牌并提取到数据库中。如果找到匹配项,它将对用户进行身份验证。
security:
# http://symfony.com/doc/current/book/security.html#where-do-users-come-from-
firewalls:
# disables authentication for assets and the profiler, adapt it according to your needs
dev:
pattern: ^/(_(profiler|wdt)|css|images|js)/
security: false
main:
stateless: true
provider: customProvider
guard:
authenticators:
- App\Security\TokenAuthenticator
src/Security/TokenAuthenticator.php
namespace App\Security;
use App\Entity\User;
use Doctrine\ORM\EntityManagerInterface;
use Symfony\Component\HttpFoundation\JsonResponse;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\Security\Core\Authentication\Token\TokenInterface;
use Symfony\Component\Security\Core\Exception\AuthenticationException;
use Symfony\Component\Security\Core\User\UserInterface;
use Symfony\Component\Security\Core\User\UserProviderInterface;
use Symfony\Component\Security\Guard\AbstractGuardAuthenticator;
class TokenAuthenticator extends AbstractGuardAuthenticator
private $em;
public function __construct(EntityManagerInterface $em)
$this->em = $em;
/**
* Called on every request to decide if this authenticator should be
* used for the request. Returning false will cause this authenticator
* to be skipped.
*/
public function supports(Request $request)
return true;
/**
* Called on every request. Return whatever credentials you want to
* be passed to getUser() as $credentials.
*/
public function getCredentials(Request $request)
return [
'token' => // recover token from cookies,
];
public function getUser($credentials, UserProviderInterface $userProvider)
$token = $credentials['token'];
if (null === $token)
return;
// if a User object, checkCredentials() is called
return $this->em->getRepository(User::class)
->findOneBy(['token' => $token]);
public function checkCredentials($credentials, UserInterface $user)
// check credentials - e.g. make sure the password is valid
// no credential check is needed in this case
// return true to cause authentication success
return true;
public function onAuthenticationSuccess(Request $request, TokenInterface $token, $providerKey)
// on success, let the request continue
return null;
public function onAuthenticationFailure(Request $request, AuthenticationException $exception)
$data = [
'message' => strtr($exception->getMessageKey(), $exception->getMessageData())
// or to translate this message
// $this->translator->trans($exception->getMessageKey(), $exception->getMessageData())
];
return new JsonResponse($data, Response::HTTP_FORBIDDEN);
/**
* Called when authentication is needed, but it's not sent
*/
public function start(Request $request, AuthenticationException $authException = null)
$data = [
// you might translate this message
'message' => 'Authentication Required'
];
return new JsonResponse($data, Response::HTTP_UNAUTHORIZED);
public function supportsRememberMe()
return false;
希望能解决你的问题
【讨论】:
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