FOSUserBundle AJAX使用Symfony2登录(路由)

Question

我正在尝试使用FOSUserBundle进行AJAX身份验证.

我创建了一个Handler目录,其中包含AuthenticationHandler class:

<?php

namespace BirdOffice\UserBundle\Handler;

use Symfony\Component\HttpFoundation\Response;
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\SecurityContextInterface;
use Symfony\Component\Security\Http\Authentication\AuthenticationSuccessHandlerInterface;
use Symfony\Component\Security\Http\Authentication\AuthenticationFailureHandlerInterface;

class AuthenticationHandler implements AuthenticationSuccessHandlerInterface, AuthenticationFailureHandlerInterface
{
    private $router;
    private $session;

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

    /**
     * onAuthenticationSuccess
     *
     * @param   Request $request
     * @param   TokenInterface $token
     * @return  Response
     */
    public function onAuthenticationSuccess( Request $request, TokenInterface $token )
    {
        // if AJAX login
        if ( $request->isXmlHttpRequest() ) {

            $array = array( 'success' => true ); // data to return via JSON
            $response = new Response( json_encode( $array ) );
            $response->headers->set( 'Content-Type', 'application/json' );

            return $response;

            // if form login
        } else {

            if ( $this->session->get('_security.main.target_path' ) ) {

                $url = $this->session->get( '_security.main.target_path' );

            } else {

                $url = $this->router->generate( 'home_page' );

            } // end if

            return new RedirectResponse( $url );

        }
    }

    /**
     * onAuthenticationFailure
     *
     * @param   Request $request
     * @param   AuthenticationException $exception
     * @return  Response
     */
    public function onAuthenticationFailure( Request $request, AuthenticationException $exception )
    {
        // if AJAX login
        if ( $request->isXmlHttpRequest() ) {

            $array = array( 'success' => false, 'message' => $exception->getMessage() ); // data to return via JSON
            $response = new Response( json_encode( $array ) );
            $response->headers->set( 'Content-Type', 'application/json' );

            return $response;

            // if form login
        } else {

            // set authentication exception to session
            $request->getSession()->set(SecurityContextInterface::AUTHENTICATION_ERROR, $exception);

            return new RedirectResponse( $this->router->generate( 'login_route' ) );
        }
    }
}

我创建了一个login Javascript function:

function login() {
    $.ajax({
        type: "POST",
        url: Routing.generate('check_login_ajax'),
        dataType: 'json',
        data: {
            _username: $('#username').val(),
            _password: $('#password').val(),
            _remember_me: false,
            _csrf_token: $('#_csrf_token').val()
        }
    }).done(function(data) {
        console.log(data);
    }).fail(function(data) {
        console.log(data);
    });
}

在我routingAjax.yml,我添加了以下行来覆盖FOSUserBundle security route:

check_login_ajax:
    pattern:  /check_login_ajax
    defaults: { _controller: FOSUserBundle:Security:check }
    requirements:
        _method:  POST
    options:
        expose: true

在我的全局security.yml文件,我已经加入了check_path,success_handler和failure_handler部分:

firewalls:
        main:
            pattern: ^/
            form_login:
                login_path: fos_user_registration_register
                check_path:      check_login_ajax
                success_handler: user.security.authentication_handler
                failure_handler: user.security.authentication_handler
                provider: fos_userbundle
                csrf_provider: form.csrf_provider
            logout:
                  path:   fos_user_security_logout
                  target: /
            anonymous:    true

我的第一个问题是:AJAX返回此消息:"无效的CSRF令牌".(但我发送了一个用PHP生成的好文章,也许我错过了这样做的东西).这是我的PHP代码:

<?php
  $csrfProvider = $this->container->get('form.csrf_provider');
  $csrfToken = $csrfProvider->generateCsrfToken('popUpUser');
?>

第二个问题:我的登录页面(不是AJAX)不再起作用,因为FOSUserBundle登录的原始路由已被覆盖.

PS:昨天我发布了一条消息:FOSUserBundle(登录/注册)+ AJAX + Symfony2,但我已经解释了我的问题.提前抱歉.

Answer 1

第一个问题:您正在发送无效的CSRF令牌.在Symfony的2.3,你可以用它生成{{ csrf_token('authenticate') }}的模板的内部input的value.

第二个问题:不要覆盖路线,只需使用原始路线:fos_user_security_check.

一般来说:如果你使用AuthenticationSuccessHandler扩展Symfony\Component\Security\Http\Authentication\DefaultAuthenticationSuccessHandler你的方法可能看起来像这样:

public function onAuthenticationSuccess(Request $request, TokenInterface $token)
{
    if ($request->isXmlHttpRequest()) {
        return new JsonResponse(array('success' => true));
    }

    return parent::onAuthenticationSuccess($request, $token);
}

为AuthenticationFailureHandler扩展做类似的事情Symfony\Component\Security\Http\Authentication\DefaultAuthenticationFailureHandler.