我是AJAX的新手。但我正在尝试了解这是如何工作的。我正在使用symfony2与fos用户捆绑,我想实现AJAX到我的登录表单。所以我这样做:
login.html.twig
<script>
$('#_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) {
$('.tab-1').prepend('<div />').html(data.message);
} else {
window.location.href = data.targetUrl;
}
}
});
</script>
<div id="tab-1" class="login_form">
<form action="{{ path("fos_user_security_check") }}" role="form" method="post">
<label for="username"><strong>User Name / Email Address</strong>
<input type="text" id="username" name="_username" value="{{ last_username }}" required="required" />
</label>
<label for="password"><strong>Password</strong>
<input type="password" id="password" name="_password" required="required" />
</label>
<label for="password"><strong>Remember Me</strong>
<input type="checkbox" id="remember_me" name="_remember_me" value="on" />
</label>
<input type="submit" class="submitBut" id="_submit" name="_submit" value="{{ 'security.login.submit'|trans({}, 'FOSUserBundle') }}" />
</form>
</div>
然后在提交时转到这个文件:-
<?php
namespace XXXX\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;
use Symfony\Component\Translation\Translator;
use Symfony\Component\Translation\MessageSelector;
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) {
$translator = new Translator('fr_FR');
//$result = array(
// 'success' => false,
// 'function' => 'onAuthenticationFailure',
// 'error' => true,
// 'message' => $this->translator->trans($exception->getMessage(), array(), 'FOSUserBundle')
//);
$result = array('success' => false);
$response = new Response(json_encode($result));
$response->headers->set('Content-Type', 'application/json');
return $response;
}
}
提交表单时,请在login_check url中显示:
{"success":false}
但我想当结果为假,然后返回相同的表单,我试图登录(我的意思是相同的弹出式div)?
我的代码ajax或动作返回有什么问题?或者我返回的是正确的?
发布于 2014-01-09 05:09:23
window.location
将重新加载整个页面。我想这不是您想要的结果,因为您正在使用AJAX ( AJAX的漏洞在于不重新加载页面),如果登录不成功,您可能会显示一条错误消息。我建议您在html表单中添加一个错误div
<div class='error' style="display:none" > ooups an erro occured </div>
然后在ajax调用中显示它或添加一个重要的消息错误:
if (data.sucess == false) {
$('.tab-1').prepend('<div />').html(data.message);
} else {
$('.error').show();
}
https://stackoverflow.com/questions/21004696
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