lee*_*eek 43 php symfony fosuserbundle
我有一个例子,我试图使用Symfony2和FOSUserBundle创建一个AJAX登录.我设置我自己success_handler和failure_handler在form_login我的security.yml文件.
这是班级:
class AjaxAuthenticationListener implements AuthenticationSuccessHandlerInterface, AuthenticationFailureHandlerInterface
{
/**
* This is called when an interactive authentication attempt succeeds. This
* is called by authentication listeners inheriting from
* AbstractAuthenticationListener.
*
* @see \Symfony\Component\Security\Http\Firewall\AbstractAuthenticationListener
* @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);
$response = new Response(json_encode($result));
$response->headers->set('Content-Type', 'application/json');
return $response;
}
}
/**
* 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, 'message' => $exception->getMessage());
$response = new Response(json_encode($result));
$response->headers->set('Content-Type', 'application/json');
return $response;
}
}
}
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这非常适合处理成功和失败的AJAX登录尝试.但是,启用时 - 我无法通过标准表单POST方法(非AJAX)登录.我收到以下错误:
Catchable Fatal Error: Argument 1 passed to Symfony\Component\HttpKernel\Event\GetResponseEvent::setResponse() must be an instance of Symfony\Component\HttpFoundation\Response, null given
我想我onAuthenticationSuccess和onAuthenticationFailure覆盖到只对XmlHttpRequests执行(AJAX请求)和简单的手工执行回原来的处理程序,如果没有.
有没有办法做到这一点?
TL; DR我希望AJAX请求登录尝试返回成功和失败的JSON响应,但我希望它不会影响通过表单POST的标准登录.
sem*_*eos 50
David的答案很好,但是对于newbs缺乏一点细节 - 所以这就是填补空白.
除了创建AuthenticationHandler之外,您还需要使用创建处理程序的包中的服务配置将其设置为服务.默认的bundle生成创建了一个xml文件,但我更喜欢yml.这是一个示例services.yml文件:
#src/Vendor/BundleName/Resources/config/services.yml
parameters:
vendor_security.authentication_handler: Vendor\BundleName\Handler\AuthenticationHandler
services:
authentication_handler:
class: %vendor_security.authentication_handler%
arguments: [@router]
tags:
- { name: 'monolog.logger', channel: 'security' }
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您需要修改DependencyInjection包扩展以使用yml而不是xml,如下所示:
#src/Vendor/BundleName/DependencyInjection/BundleExtension.php
$loader = new Loader\YamlFileLoader($container, new FileLocator(__DIR__.'/../Resources/config'));
$loader->load('services.yml');
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然后在您的应用程序的安全配置中,设置对刚刚定义的authentication_handler服务的引用:
# app/config/security.yml
security:
firewalls:
secured_area:
pattern: ^/
anonymous: ~
form_login:
login_path: /login
check_path: /login_check
success_handler: authentication_handler
failure_handler: authentication_handler
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Dav*_*les 31
namespace YourVendor\UserBundle\Handler;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\HttpFoundation\RedirectResponse;
use Symfony\Bundle\FrameworkBundle\Routing\Router;
use Symfony\Component\Security\Core\Authentication\Token\TokenInterface;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\Security\Http\Authentication\AuthenticationSuccessHandlerInterface;
use Symfony\Component\Security\Http\Authentication\AuthenticationFailureHandlerInterface;
use Symfony\Component\Security\Core\Exception\AuthenticationException;
class AuthenticationHandler
implements AuthenticationSuccessHandlerInterface,
AuthenticationFailureHandlerInterface
{
private $router;
public function __construct(Router $router)
{
$this->router = $router;
}
public function onAuthenticationSuccess(Request $request, TokenInterface $token)
{
if ($request->isXmlHttpRequest()) {
// Handle XHR here
} 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.target_path')) {
$url = $targetPath;
} else {
// Otherwise, redirect him to wherever you want
$url = $this->router->generate('user_view', array(
'nickname' => $token->getUser()->getNickname()
));
}
return new RedirectResponse($url);
}
}
public function onAuthenticationFailure(Request $request, AuthenticationException $exception)
{
if ($request->isXmlHttpRequest()) {
// Handle XHR here
} else {
// Create a flash message with the authentication error message
$request->getSession()->setFlash('error', $exception->getMessage());
$url = $this->router->generate('user_login');
return new RedirectResponse($url);
}
}
}
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