php – Symfony 3.2 FOSUserBundle Ajax登录

在FOSUserBundle中,我想在用户登录后重定向用户而不加载页面(AJAX查询)到fos_user_profile_show路由.我坚持到这一点.论坛中有类似的主题,但它们已经过时了.

AuthenticationHandler.php

<?php

namespace AppBundle\Handler;

use Symfony\Component\HttpFoundation\JsonResponse;
use Symfony\Component\HttpFoundation\RedirectResponse;
use Symfony\Component\Routing\RouterInterface;
use Symfony\Component\HttpFoundation\Session\Session;
use Symfony\Component\Security\Core\Authentication\Token\TokenInterface;
use Symfony\Component\Security\Core\Exception\AuthenticationException;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\Security\Core\Security;
use Symfony\Component\Security\Http\Authentication\AuthenticationSuccessHandlerInterface;
use Symfony\Component\Security\Http\Authentication\AuthenticationFailureHandlerInterface;

/**
 * Class AuthenticationHandler
 * @package AppBundle\Handler
 */
class AuthenticationHandler implements AuthenticationSuccessHandlerInterface, AuthenticationFailureHandlerInterface
{
    /**
     * @var RouterInterface
     */
    private $router;
    /**
     * @var Session
     */
    private $session;


    /**
     * AuthenticationHandler constructor.
     * @param RouterInterface $router
     * @param Session $session
     */
    public function __construct(RouterInterface $router, Session $session)
    {
        $this->router = $router;
        $this->session = $session;
    }

    /**
     * @param Request $request
     * @param TokenInterface $token
     * @return JsonResponse|RedirectResponse
     */
    public function onAuthenticationSuccess(Request $request, TokenInterface $token)
    {

        if ($request->isXmlHttpRequest()) {
            return new JsonResponse(array('success' => true));
        }
        else {
            $url = $this->router->generate('fos_user_profile_show');
            return new RedirectResponse($url);
        }

    }

    /**
     * @param Request $request
     * @param AuthenticationException $exception
     * @return JsonResponse|RedirectResponse
     */
    public function onAuthenticationFailure(Request $request, AuthenticationException $exception)
    {
        if ($request->isXmlHttpRequest()) {
            return new JsonResponse(array('success' => false, 'message' => $exception->getMessage()));
        } else {
            $request->get('session')->set(Security::AUTHENTICATION_ERROR, $exception);
            return new RedirectResponse($this->router->generate('fos_user_security_login'));
        }
    }
}

services.yml

app.security.authentication_handler:
    class: AppBundle\Handler\AuthenticationHandler
    public: false
    arguments:
            - "@router"
            - "@session"

security.yml

firewalls:
    main:
        pattern: ^/
        form_login:
            provider: fos_userbundle
            csrf_token_generator: security.csrf.token_manager
            check_path: fos_user_security_check
            success_handler: app.security.authentication_handler
            failure_handler: app.security.authentication_handler
        logout:       true
        anonymous:    true

login_content.html.twig

<script>
    $(document).ready(function(){

        $('#_submit').click(function(e){
            e.preventDefault();
            $.ajax({
                type        : $('form').attr( 'method' ),
                url         : $('form').attr( 'action' ),
                data        : $('form').serialize(),
                success     : function(data, status, object) {
                    if (data.success == false) {
                        console.log(data.message);
                    } else {
                        window.location.href = data.targetUrl;
                    }
                }
            });

    });
</script>

解决方法:

我修好了这个;

$url = $this->router->generate('fos_user_profile_show');
return new JsonResponse(array('success' => true));

顺便说一下,感谢所有这些代码.

标签：php,ajax,fosuserbundle,symfony
来源： https://codeday.me/bug/20190705/1391517.html