小编Val*_*tin的帖子

有没有人已经使用Symfony 2和FOS User Bundle在Bootstrap模式中构建了一个登录表单？

这就是我现在拥有的:

SRC/Webibli/UserBundle /资源/配置/ service.yml

authentication_handler:
    class:        Webibli\UserBundle\Handler\AuthenticationHandler
    arguments:    [@router, @security.context, @fos_user.user_manager, @service_container]

应用程序/配置/ security.yml

form_login:
    provider: fos_userbundle
    success_handler: authentication_handler
    failure_handler: authentication_handler

SRC/Webibli/UserBundle /处理器/ AuthenticationHandler.php

<?php

namespace Webibli\UserBundle\Handler;

use Symfony\Component\Security\Http\Authentication\AuthenticationFailureHandlerInterface;
use Symfony\Component\Security\Http\Authentication\AuthenticationSuccessHandlerInterface;
use Symfony\Component\Security\Core\Authentication\Token\TokenInterface;
use Symfony\Component\Routing\RouterInterface;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\HttpFoundation\RedirectResponse;
use Symfony\Component\Routing\Router;
use Symfony\Component\Security\Core\SecurityContext;
use Symfony\Component\Security\Core\Exception\AuthenticationException;


class AuthenticationHandler implements AuthenticationSuccessHandlerInterface, AuthenticationFailureHandlerInterface
{

    protected $router;
    protected $security;
    protected $userManager;
    protected $service_container;

    public function __construct(RouterInterface $router, SecurityContext $security, $userManager, $service_container)
    {
        $this->router = $router;
        $this->security = $security; …