11

是否有人已经使用 Symfony 2 和 FOS User Bundle 在 Bootstrap 模式中构建了登录表单？

这是我现在拥有的：

src/Webibli/UserBundle/Resources/config/service.yml 

authentication_handler:
    class:        Webibli\UserBundle\Handler\AuthenticationHandler
    arguments:    [@router, @security.context, @fos_user.user_manager, @service_container]

应用程序/配置/security.yml

form_login:
    provider: fos_userbundle
    success_handler: authentication_handler
    failure_handler: authentication_handler

src/Webibli/UserBundle/Handler/AuthenticationHandler.php

<?php

namespace Webibli\UserBundle\Handler;

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


class AuthenticationHandler implements AuthenticationSuccessHandlerInterface, AuthenticationFailureHandlerInterface
{

    protected $router;
    protected $security;
    protected $userManager;
    protected $service_container;

    public function __construct(RouterInterface $router, SecurityContext $security, $userManager, $service_container)
    {
        $this->router = $router;
        $this->security = $security;
        $this->userManager = $userManager;
        $this->service_container = $service_container;

    }
    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;
        }
        else {
            // Create a flash message with the authentication error message
            $request->getSession()->getFlashBag()->set('error', $exception->getMessage());
            $url = $this->router->generate('fos_user_security_login');

            return new RedirectResponse($url);
        }

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

这是我正在加载到 Bootstrap 模式中的 Twig 视图：

{% extends 'UserBundle::layout.html.twig' %}
{% trans_default_domain 'FOSUserBundle' %}
{% block user_content %}
<script>
    $('#_submit').click(function(e){
        e.preventDefault();
        $.ajax({
            type        : $('form').attr( 'method' ),
            url         : $('form').attr( 'action' ),
            data        : $('form').serialize(),
            success     : function(data, status, object) {
                console.log( status );
                console.log( object.responseText );
            }
    });
});
</script>
<div class="modal-dialog">
    <div class="modal-content">
        <form action="{{ path("fos_user_security_check") }}" method="post" role="form" data-async data-target="#rating-modal" class="text-left">
        <div class="modal-header">
            <button type="button" class="close" data-dismiss="modal" aria-hidden="true">&times;</button>
            <h4 class="modal-title">{{ 'layout.login'|trans }}</h4>
        </div>
        <div class="modal-body">
            {% if error %}
                <div>{{ error|trans }}</div>
            {% endif %}
            <input type="hidden" name="_csrf_token" value="{{ csrf_token }}" />
            <div class="form-group container">
                <label for="email">{{ 'security.login.username_email'|trans }}</label>
                <input type="text" class="form-control" id="username" name="_username" value="{{ last_username }}" required="required" placeholder="adresse@email.com">
            </div>
            <div class="form-group container">
                <label for="password">{{ 'security.login.password'|trans }}</label><br />
                <input type="password" id="password" name="_password" required="required" class="form-control" placeholder="********">
            </div>
            <div class="form-group container">
                <label for="remember_me">
                    <input type="checkbox" id="remember_me" name="_remember_me" value="on" />
                    {{ 'security.login.remember_me'|trans }}
                </label>
            </div>
        </div>
        <div class="modal-footer">
          <input type="submit" id="_submit" name="_submit" value="{{ 'security.login.submit'|trans }}" class="btn btn-primary">
        </div>
    </form>
</div>
</div>
{% endblock %}

登录表单在没有 AJAX 的情况下工作得很好。如果出现问题，我只是想在模式中的表单上出现错误，或者如果登录成功则重定向用户。

谁能解释如何实现这一目标？

4

2 回答 2

8

我找到了解决方案。这是我添加到我的 javascript 中的内容，

<script>
    $(document).ready(function(){
        $('#_submit').click(function(e){
            e.preventDefault();
            $.ajax({
                type        : $('form').attr( 'method' ),
                url         : '{{ path("fos_user_security_check") }}',
                data        : $('form').serialize(),
                dataType    : "json",
                success     : function(data, status, object) {
                    if(data.error) $('.error').html(data.message);
                },
                error: function(data, status, object){
                    console.log(data.message);
                }
            });
        });
    });
</script>

这是我onAuthenticationFailure的处理程序的方法，

public function onAuthenticationFailure(Request $request, AuthenticationException $exception) {
    $result = array(
        'success' => false, 
        'function' => 'onAuthenticationFailure', 
        'error' => true, 
        'message' => $this->translator->trans($exception->getMessage(), array(), 'FOSUserBundle')
    );
    $response = new Response(json_encode($result));
    $response->headers->set('Content-Type', 'application/json');

    return $response;
}

我认为我的 Ajax 方法中的 URL 是错误的。谢谢你的建议。

于 2013-10-19T07:24:13.540 回答
6

我猜你要找的是这个：Symfony2 ajax login。

你的 javascript 看起来会…… 像这样：

$('#_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.sucess == false) {
                    $('.modal-body').prepend('<div />').html(data.message);
                } else {
                    window.location.href = data.targetUrl;
                }
            }
    });

您还必须修改 onAuthenticationSuccess-Method 的 isXmlHttpRequest 部分：

[...]
if ($request->isXmlHttpRequest()) {
            $targetUrl = $request->getSession()->get('_security.target_path');
            $result = array('success' => true, 'targetUrl' => targetUrl );
            $response = new Response(json_encode($result));
            $response->headers->set('Content-Type', 'application/json');
            return $response;
        }
[...]
于 2013-10-18T19:37:05.703 回答