Symfony2 AJAX登录

Question

我有一个示例，尝试使用Symfony2和FOSUserBundle创建AJAX登录。我设置我自己success_handler和failure_handler在form_login我的security.yml文件。

这是课程：

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登录尝试都非常有用。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

我希望自己的重写onAuthenticationSuccess和onAuthenticationFailure重写仅对XmlHttpRequests（AJAX请求）执行，如果没有，则将执行交还给原始处理程序。

有没有办法做到这一点？

TL; DR我希望AJAX请求登录尝试返回JSON响应以表示成功和失败，但我希望它不影响通过POST表单进行的标准登录。

Answer 1

David的回答很好，但是对于新手来说，它缺少一些细节-
因此，这是要填补空白。

除了创建AuthenticationHandler之外，还需要使用创建处理程序的捆绑软件中的服务配置将其设置为服务。默认的捆绑软件生成将创建一个xml文件，但我更喜欢yml。这是一个示例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');

然后在应用程序的安全性配置中，设置对您刚定义的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