登录失败后如何返回引荐来源网址? [英] How to go back to referer after login failure?

问题描述

为成功登录，有一个参数use_referer: true.对于登录失败，只有failure_path，这不是我想要的.

For login success there is a parameter use_referer: true. For login failure there is only failure_path, which isn't what I'm looking for.

第二件事:如何做到这一点并传递错误消息?

Second thing: How to do that and pass error message?

第三件事:注销后如何返回引荐来源网址?

Third thing: How to go back to referrer after logout?

推荐答案

我解决了.

有解决方案:如何在Symfony 2中的login_check之后禁用重定向

这是解决我的问题的代码:

and here is code which solves my problem:

<?php

namespace Acme\MainBundle\Handler;

use Symfony\Component\Security\Http\Logout\LogoutSuccessHandlerInterface;
use Symfony\Component\Security\Http\Authentication\AuthenticationFailureHandlerInterface;
use Symfony\Component\Security\Core\Exception\AuthenticationException;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\RedirectResponse;

class AuthenticationHandler implements AuthenticationFailureHandlerInterface, LogoutSuccessHandlerInterface
{
    public function onAuthenticationFailure(Request $request, AuthenticationException $exception)
    {       
        $referer = $request->headers->get('referer');       
        $request->getSession()->setFlash('error', $exception->getMessage());

        return new RedirectResponse($referer);
    }

    public function onLogoutSuccess(Request $request) 
    {
        $referer = $request->headers->get('referer');
        return new RedirectResponse($referer);
    }
}

要处理事件，请添加到security.yml中，例如:

to handle events add to security.yml for example:

form_login:
    check_path: /access/login_check
    login_path: /
    use_referer: true
    failure_handler: authentication_handler  
logout:
    path:   /access/logout
    target: /
    success_handler: authentication_handler 

并转到config.yml:

and to config.yml:

services:
    authentication_handler:
        class: Acme\MainBundle\Handler\AuthenticationHandler

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