1

我正在使用带有 FOSUserBundle 的 Symfony2，并且我想使用 Ajax 进行用户登录。我使用扩展 DefaultAuthenticationFailureHandler 类的服务：

<?php
namespace Marquis\WebsiteBundle\Handler;

use Psr\Log\LoggerInterface;
use Symfony\Component\Translation\Translator;
use Symfony\Component\HttpFoundation\JsonResponse;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\HttpFoundation\RedirectResponse;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpKernel\HttpKernelInterface;
use Symfony\Component\Security\Core\Exception\AuthenticationException;
use Symfony\Component\Security\Http\Authentication\DefaultAuthenticationFailureHandler;
use Symfony\Component\Security\Http\HttpUtils;

class AuthenticationFailureHandler extends DefaultAuthenticationFailureHandler {

    protected $translator;

    public function __construct(HttpKernelInterface $httpKernel, HttpUtils $httpUtils, array $options, LoggerInterface $logger = null, Translator $translator = null) {
        parent::__construct($httpKernel, $httpUtils, $options, $logger);

        $this->translator = $translator;
    }

    public function onAuthenticationFailure(Request $request, AuthenticationException $exception) {
        if($request->isXmlHttpRequest()) {
            /* @var $formResponse RedirectResponse */
            $formResponse = parent::onAuthenticationFailure($request, $exception);
            $response = new JsonResponse(array('success' => false,
                'message' => $this->translator->trans($exception->getMessage(), array(), 'FOSUserBundle'),
                'form' => $formResponse->getTargetUrl()));
        } else {
            $response = parent::onAuthenticationFailure($request, $exception);
        }

        return $response;
    }
}

它工作得很好，但我想将渲染视图转换为“表单”JsonResponse 值。

是否可以从 RedirectResponse 对象生成视图？

4

1 回答 1

0

如果我理解正确，您想呈现一个视图并在登录或注册页面上显示该视图对吗？

如果是这样，您可以直接返回您的视图，并从 javascript 代码中解析为 html 并附加到您想要的任何位置。

return $this->twig->render("SampleBundle:sample.html.twig", $data); //$data is not required

你可以像这样将加载的html附加到你的容器中；

$.ajax({
  dataType:"html",
  success: function(response){
    response.append('#container');
  }
});
于 2013-09-14T15:03:32.170 回答