有我的问题，我需要分几步提交表格。我有应用表格和登录表单
<form id="app_form" action="{{ path('app_create') }}" method="post" {{ form_enctype(formApp) }}>
<div class="row-fluid">
<div class="span2">{{ form_label(formApp.name, 'Name'|trans) }}</div>
<div class="span4">{{ form_widget(formApp.name, { required : true }) }}</div>
<div class="span2" id="error_app_name">{{ form_errors(formApp.name) }}</div>
</div>
<div class="row-fluid">
<div class="span2">{{ form_label(formApp.description, 'Description'|trans) }}</div>
<div class="span4">{{ form_widget(formApp.description, { required : true }) }}</div>
<div class="span2">{{ form_errors(formApp.description) }}</div>
</div>
<div class="row-fluid">
<div class="span2">{{ form_label(formApp.iosUrl, 'iOS'|trans) }}</div>
<div class="span4">{{ form_widget(formApp.iosUrl) }}</div>
<div class="span2">{{ form_errors(formApp.iosUrl) }}</div>
</div>
<div class="row-fluid">
<div class="span2">{{ form_label(formApp.androidBundle, 'Android Bundle'|trans) }}</div>
<div class="span4">{{ form_widget(formApp.androidBundle) }}</div>
<div class="span2">{{ form_errors(formApp.androidBundle) }}</div>
</div>
<div class="row-fluid">
<div class="span2">{{ form_label(formApp.wpUrl, 'Windows Phone'|trans) }}</div>
<div class="span4">{{ form_widget(formApp.wpUrl) }}</div>
<div class="span2">{{ form_errors(formApp.wpUrl) }}</div>
</div>
<div class="row-fluid">
<div class="span2">{{ form_label(formApp.bbUrl, 'Blackberry'|trans) }}</div>
<div class="span4">{{ form_widget(formApp.bbUrl) }}</div>
<div class="span2">{{ form_errors(formApp.bbUrl) }}</div>
</div>
<div class="row-fluid">
<div class="span2">{{ form_label(formApp.fallbackUrl, 'Fallback Url'|trans) }}</div>
<div class="span4">{{ form_widget(formApp.fallbackUrl) }}</div>
<div class="span2">{{ form_errors(formApp.fallbackUrl) }}</div>
</div>
{{ form_rest(formApp) }}
<button type="submit" class="btn">{{ 'Next step'|trans }}</button>
</form>
<form id="form_login">
<input type="hidden" id="login_csrf_token" name="_csrf_token" value="{{ csrf_token }}" />
<label for="login_username">{{ 'security.login.username'|trans({}, 'FOSUserBundle') }}</label>
<input type="text" id="login_username" name="_username" value="{{ last_username }}" required="true" />
<label for="login_password">{{ 'security.login.password'|trans({}, 'FOSUserBundle') }}</label>
<input type="password" id="login_password" name="_password" required="true" />
<input type="checkbox" id="login_remember_me" name="_remember_me" value="on" />
<label for="login_remember_me">{{ 'security.login.remember_me'|trans({}, 'FOSUserBundle') }}</label>
<button type="submit" class="btn">{{ 'Login'|trans }}</button>
<button class="btn" id="register_show" type="button">{{ 'Want to register ?'|trans }}</button>
<div id="error_login"></div>
</form>
我想提交登录表单en ajax然后提交应用程序表单，但是当我这样做时我有错误
CSRF令牌无效。请尝试重新提交表单
Javascript代码：
$('#form_login').submit(function(){
event.preventDefault();
$.post('./login_check', { _csrf_token : $('#login_csrf_token').val() ,
_username : $('#login_username').val() ,
_password : $('#login_password').val(),
_remember_me : $('#login_remember_me').val() },
function (data) {
console.log(data);
/**
* If the login was ok we submit the app form
*/
if(data.success){
$('#app_form').submit();
}
/**
* Else we sow to the user the error
*/
else{
$('#error_login').html(data.message);
}
}
);
});
登录表单返回成功，但是当我提交应用程序表单 $（'＃app_form'）。submit（）; 时，错误会显示在下一页。
提前致谢：）
答案 0 :(得分：3)
我选择的解决方案是在每个LoginAjax操作中重新生成令牌，并在响应中发送它（以JSON格式化）。然后我只需要更新所有具有令牌字段的表单（可以通过为每个令牌设置特定类来使用。
实现AuthenticationSuccessHandler的第一步 定义为此
acme.security.ajax_handler:
class: Acme\UserBundle\Listener\AjaxAuthenticationListener
arguments: ["@form.csrf_provider", "@router"]
AjaxAuthenticationListener类
<?php
namespace Travelyo\UserBundle\Listener;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\HttpFoundation\RedirectResponse;
use Symfony\Component\Security\Core\Authentication\Token\TokenInterface;
use Symfony\Component\Security\Core\Exception\AuthenticationException;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\Security\Http\Authentication\AuthenticationFailureHandlerInterface;
use Symfony\Component\Security\Http\Authentication\AuthenticationSuccessHandlerInterface;
/**
* Refere to http://stackoverflow.com/questions/8607212/symfony2-ajax-login
* @author yoni
*
*/
class AjaxAuthenticationListener implements AuthenticationSuccessHandlerInterface, AuthenticationFailureHandlerInterface {
protected $csrf_provider;
protected $router;
/**
* In case we have Failure I need to provide a new csrf token
* @param unknown_type $csrf_provider
* @author Yoni Alhadeff
*/
public function __construct($csrf_provider, $router)
{
$this->csrf_provider = $csrf_provider;
$this->router = $router;
}
/**
* This is called when an interactive authentication attempt succeeds. This
* is called by authentication listeners inheriting from
* AbstractAuthenticationListener.
*
* @see SymfonyComponentSecurityHttpFirewallAbstractAuthenticationListener
* @param Request $request
* @param TokenInterface $token
* @return Response the response to return
*/
public function onAuthenticationSuccess(Request $request, TokenInterface $token)
{
if ($request->isXmlHttpRequest())
{
$result = array('success' => true, 'token'=>$this->csrf_provider->generateCsrfToken('unknown'));
$response = new Response(json_encode($result));
$response->headers->set('Content-Type', 'application/json');
return $response;
} else
{
// If the user tried to access a protected resource and was forces to login
// redirect him back to that resource
if ($targetPath = $request->getSession()->get('_security.account.target_path'))
{
$url = $targetPath;
} else
{
// Otherwise, redirect him to wherever you want
$url = $this->router->generate('homepage_route_name', array());
}
return new RedirectResponse($url);
}
}
/**
* 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())
{
$result = array('success' => false, 'error'=>$exception->getMessage(),'token'=>$this->csrf_provider->generateCsrfToken('authenticate'));
$response = new Response(json_encode($result));
$response->headers->set('Content-Type', 'application/json');
return $response;
} else
{
// Create a flash message with the authentication error message
$request->getSession()->setFlash('error', $exception->getMessage());
$url = $this->router->generate('fos_frontend_security_login');
return new RedirectResponse($url);
}
}
}
失败与失败使用AJAX时成功登录，我在响应中添加新令牌。
在javascript中的病房之后，在AJAX调用的Succes：function（）中，我只需要更新字段
类似的东西。
var route = $('#loginUrl').val();
var data= '_username=' + $("#username").val()+"&_password="+$('#password').val()+"&_csrf_token="+$('#csrf_token').val();
$.ajax( {
type : "POST",
url : route,
data: data,
dataType: 'json',
cache : false,
success : function(response, value){
//Response.success came from the JSON defined in the AjaxAuthenticationListener, he's not linked to the AJAX call
if(response.success==true)
{
}
$('#csrf_token').val(response.token); //Token has to be regenerate
$(".tokenField").val(response.token);
}
});
我认为这是更容易的解决方案。
答案 1 :(得分：0)
CSRF令牌部分使用cookie。当您在一个页面上生成两个CSRF并提交其中一个表单时，您将使cookie无效。
如果没有对框架本身进行一些扩展，我只能真正看到一种解决方法 - 而且它相当迂回：
您可以做的是设置一个生成您的应用表单的控制器。
在初始页面加载时，您的控制器将加载登录表单和应用程序表单。在通过AJAX提交登录表单后，您还将仅请求应用程序表单的控制器（这也将为用户提供新的cookie）。使用javascript，您可以从新表单中提取NEW csrf标记并将其注入原始应用程序表单。然后，当您提交应用程序表单时，它应该有一个新的，有效的csrf令牌。
举例说明：
获取应用表单和登录表单 - ＆gt;通过AJAX提交登录 - ＆gt;在后台通过AJAX获取应用程序表单 - ＆gt;窃取新的应用程序表单的csrf令牌并将其注入第一个应用程序表单 - ＆gt;提交申请表。
答案 2 :(得分：0)
使用$('form').serialize()：
$('#form_login').submit(function(){
$.ajax({
url: $(this).attr('action'),
method: $(this).attr('method'),
data: $(this).serialize(),
success: function(response){
alert(response);
}
});
});
在控制器使用中：
return new Response('ok')以获得良好的回应并return new Response('bad :(', 500)回复错误