Symfony手动登录用户
存在一个创建用户实体的页面(这超出了正常的注册流程) 创建用户时,他们应该登录,guardHandler与验证器一起使用,如下所示Symfony手动登录用户,symfony,symfony4,fosuserbundle,Symfony,Symfony4,Fosuserbundle,存在一个创建用户实体的页面(这超出了正常的注册流程) 创建用户时,他们应该登录,guardHandler与验证器一起使用,如下所示 use App\Security\FakeAuthenticator; use Symfony\Component\Security\Guard\GuardAuthenticatorHandler; $response = $guardHandler->authenticateUserAndHandleSuccess(
use App\Security\FakeAuthenticator;
use Symfony\Component\Security\Guard\GuardAuthenticatorHandler;
$response = $guardHandler->authenticateUserAndHandleSuccess(
$user, // the User object you just created
$request,
$authenticator, // authenticator whose onAuthenticationSuccess you want to use
'main' // the name of your firewall in security.yaml
);
但是,验证器很混乱,它只为AuthenticationSuccess上的一个方法创建
use Symfony\Component\Security\Guard\AbstractGuardAuthenticator;
class FakeAuthenticator extends AbstractGuardAuthenticator
{
public function supports(Request $request)
{
return false;
}
public function getCredentials(Request $request)
{
throw new \RuntimeException('Unreachable code');
}
public function getUser($credentials, UserProviderInterface $userProvider)
{
throw new \RuntimeException('Unreachable code');
}
public function checkCredentials($credentials, UserInterface $user)
{
throw new \RuntimeException('Unreachable code');
}
public function onAuthenticationSuccess(Request $request, TokenInterface $token, $providerKey)
{
return null;
}
public function onAuthenticationFailure(Request $request, AuthenticationException $exception)
{
throw new \RuntimeException('Unreachable code');
}
public function start(Request $request, AuthenticationException $authException = null)
{
throw new \RuntimeException('Unreachable code');
}
public function supportsRememberMe()
{
return true;
}
}
必须实现很多方法,因为方法handleAuthenticationSuccess
需要一个实现AuthenticatorInterface
的类
代码正常,用户已登录,但感觉它不是最干净的解决方案,是否有其他方法登录用户
FosUserBundle正在项目中使用,下面的方法确实有效,但我不确定是否支持在loginManager上调用方法,我在文档中找不到任何内容,我不希望我的代码依赖于可能更改的功能
\FOS\UserBundle\Security\LoginManagerInterface::logInUser('main', $user, $response);
我决定使用
loginManager
及其公共方法logInUser
,这是最干净的解决方案,无需为单个方法创建额外的类
use FOS\UserBundle\Security\LoginManager;
...
public function createUserInControllerAction(LoginManagerInterface $loginManager): Response
{
$user = new User(); // create user however you like
$loginManager->logInUser('main', $user, $response);
return $this->json(['success']);
}