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。