1
我是AJAX新手。但我正在努力學習這是如何工作的。 我正在使用symfony2與fos用戶包，我想實現AJAX到我的登錄表單。 所以我這樣做：如何AJAX工作
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>
，並提交時，然後去這個文件： -
<?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;
}
}
當提交表單，然後告訴我在login_check網址：
{"success":false}
但我想當結果錯誤，然後返回相同的形式，我在哪裏試試登錄（我的意思是相同的彈出div）？
我的代碼ajax或操作返回有什麼問題？ 或者我回復正確？