Symfony2 AJAX Login

2022-08-30 11:41:02

我有一个例子,我试图使用Symfony2和FOSUserBundle创建AJAX登录名。我正在我的文件中设置自己的和下面的。success_handlerfailure_handlerform_loginsecurity.yml

这是类:

class AjaxAuthenticationListener implements AuthenticationSuccessHandlerInterface, AuthenticationFailureHandlerInterface
{  
    /**
     * This is called when an interactive authentication attempt succeeds. This
     * is called by authentication listeners inheriting from
     * AbstractAuthenticationListener.
     *
     * @see \Symfony\Component\Security\Http\Firewall\AbstractAuthenticationListener
     * @param Request        $request
     * @param TokenInterface $token
     * @return Response the response to return
     */
    public function onAuthenticationSuccess(Request $request, TokenInterface $token)
    {
        if ($request->isXmlHttpRequest()) {
            $result = array('success' => true);
            $response = new Response(json_encode($result));
            $response->headers->set('Content-Type', 'application/json');
            return $response;
        }
    }

    /**
     * This is called when an interactive authentication attempt fails. This is
     * called by authentication listeners inheriting from
     * AbstractAuthenticationListener.
     *
     * @param Request                 $request
     * @param AuthenticationException $exception    
     * @return Response the response to return
     */
    public function onAuthenticationFailure(Request $request, AuthenticationException $exception)
    {
        if ($request->isXmlHttpRequest()) {
            $result = array('success' => false, 'message' => $exception->getMessage());
            $response = new Response(json_encode($result));
            $response->headers->set('Content-Type', 'application/json');
            return $response;
        }
    }
}

这对于处理成功和失败的 AJAX 登录尝试非常有用。但是,启用后 - 我无法通过标准表单POST方法(非AJAX)登录。我收到以下错误:

Catchable Fatal Error: Argument 1 passed to Symfony\Component\HttpKernel\Event\GetResponseEvent::setResponse() must be an instance of Symfony\Component\HttpFoundation\Response, null given

我希望我的和重写仅针对 XmlHttpRequests(AJAX 请求)执行,如果不是,则简单地将执行交还给原始处理程序。onAuthenticationSuccessonAuthenticationFailure

有没有办法做到这一点?

TL;DR 我希望AJAX请求的登录尝试返回JSON响应以确认成功和失败,但我希望它不会影响通过表单POST进行的标准登录。


答案 1

David的答案很好,但对于新手来说,它缺乏一点细节 - 所以这是为了填补空白。

除了创建身份验证处理程序之外,还需要使用创建处理程序的捆绑包中的服务配置将其设置为服务。默认的捆绑包生成会创建一个 xml 文件,但我更喜欢 yml。下面是一个示例 services.yml 文件:

#src/Vendor/BundleName/Resources/config/services.yml

parameters:
    vendor_security.authentication_handler: Vendor\BundleName\Handler\AuthenticationHandler

services:
    authentication_handler:
        class:  %vendor_security.authentication_handler%
        arguments:  [@router]
        tags:
            - { name: 'monolog.logger', channel: 'security' }

您需要修改 DependencyInjection 捆绑包扩展以使用 yml 而不是 xml,如下所示:

#src/Vendor/BundleName/DependencyInjection/BundleExtension.php

$loader = new Loader\YamlFileLoader($container, new FileLocator(__DIR__.'/../Resources/config'));
$loader->load('services.yml');

然后,在应用的安全配置中,设置对刚定义的authentication_handler服务的引用:

# app/config/security.yml

security:
    firewalls:
        secured_area:
            pattern:    ^/
            anonymous: ~
            form_login:
                login_path:  /login
                check_path:  /login_check
                success_handler: authentication_handler
                failure_handler: authentication_handler

答案 2
namespace YourVendor\UserBundle\Handler;

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

class AuthenticationHandler
implements AuthenticationSuccessHandlerInterface,
           AuthenticationFailureHandlerInterface
{
    private $router;

    public function __construct(Router $router)
    {
        $this->router = $router;
    }

    public function onAuthenticationSuccess(Request $request, TokenInterface $token)
    {
        if ($request->isXmlHttpRequest()) {
            // Handle XHR here
        } else {
            // If the user tried to access a protected resource and was forces to login
            // redirect him back to that resource
            if ($targetPath = $request->getSession()->get('_security.target_path')) {
                $url = $targetPath;
            } else {
                // Otherwise, redirect him to wherever you want
                $url = $this->router->generate('user_view', array(
                    'nickname' => $token->getUser()->getNickname()
                ));
            }

            return new RedirectResponse($url);
        }
    }

    public function onAuthenticationFailure(Request $request, AuthenticationException $exception)
    {
        if ($request->isXmlHttpRequest()) {
            // Handle XHR here
        } else {
            // Create a flash message with the authentication error message
            $request->getSession()->setFlash('error', $exception->getMessage());
            $url = $this->router->generate('user_login');

            return new RedirectResponse($url);
        }
    }
}

推荐