当覆盖FOSUserBundle重置密码控制器时,有一个函数调用“authenticateUser”方法(第104行):
....
$this->authenticateUser($user);
....
我的问题是我已经覆盖了Symfony身份验证处理程序,并在用户登录时拥有自己的逻辑。
修改 这是我的身份验证处理程序:
<?php
/* ... all includes ... */
class AuthenticationHandler implements AuthenticationSuccessHandlerInterface, LogoutSuccessHandlerInterface
{
private $router;
private $container;
public function __construct(Router $router, ContainerInterface $container)
{
$this->router = $router;
$this->container = $container;
}
public function onAuthenticationSuccess(Request $request, TokenInterface $token)
{
// retrieve user and session id
$user = $token->getUser();
/* ... here I do things in database when logging in, and dont want to write it again and again ... */
// prepare redirection URL
if($targetPath = $request->getSession()->get('_security.target_path')) {
$url = $targetPath;
}
else {
$url = $this->router->generate('my_route');
}
return new RedirectResponse($url);
}
}
那么,我如何从ResettingController中的身份验证处理程序调用“onAuthenticationSuccess”方法? 为了避免重写相同的代码...
感谢您的帮助!
斯坦
答案 0 :(得分:1)
您应该调用onAuthenticationSuccess方法将其作为服务加载。在你的config.yml:
authentication_handler:
class: Acme\Bundle\Service\AuthenticationHandler
arguments:
container: "@service_container"
然后,在authenticateUser函数中调用它:
protected function authenticateUser(UserInterface $user) {
try {
$this->container->get('fos_user.user_checker')->checkPostAuth($user);
} catch (AccountStatusException $e) {
// Don't authenticate locked, disabled or expired users
return;
}
$providerKey = $this->container->getParameter('fos_user.firewall_name');
$token = new UsernamePasswordToken($user, null, $providerKey, $user->getRoles());
$this->container->get('security.context')->setToken($token);
$request = $this->container->get('request');
$this->container->get('authentication_handler')->onAuthenticationSuccess($request, $token);
}
这样做可以通过您的自定义身份验证处理程序。 More info