Symfony2中的RequestListener

我正在使用symfony2.3。 我创建RequestListener并在用户登录时写入，然后执行返回主页操作 获得了一些奖励积分。这很有效。 但我如何才能做到这一点，只有当用户第一次回家行动时才能获得积分。如果用户回家行动超过一次，则不更新奖励积分

RequestListner.php

<?php

namespace XXX\ABCBundle\EventListener;

use Symfony\Component\HttpKernel\Event\GetResponseEvent;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\DependencyInjection\ContainerInterface;
use Symfony\Component\Security\Core\SecurityContextInterface;
use Symfony\Component\HttpKernel\HttpKernel;
use Doctrine\ORM\EntityManager;
use Thinkdigit\UserBundle\Entity\User;


class RequestListener
{
/**
 * @var \Symfony\Component\Security\Core\SecurityContextInterface
 */
protected $securityContext;

protected $em;

/**
 * Container
 *
 * @var ContainerInterface
 */
protected $container;

/**
 * Listener constructor
 *
 * @param ContainerInterface $container
 */
public function __construct(ContainerInterface $container, SecurityContextInterface $securityContext, EntityManager $em)
{
    $this->container = $container;
    $this->securityContext = $securityContext;
    $this->em = $em;
}

/**
 * kernel.request Event
 *
 * @param GetResponseEvent $event
 */
public function onKernelRequest(GetResponseEvent $event)
{
    $request  = $event->getRequest();
    $controller = $request->get('_controller');


$home     = array('XXX\ABCBundle\Controller\PageHomeController::indexAction', 'XXX\ABCBundle\Controller\RegistrationController::confirmedAction');


  //print_r($controller);die;
    if ( $this->container->get('security.context')->isGranted('IS_AUTHENTICATED_FULLY'))
    {
        if(in_array($controller,$home)){
            $user = $this->container->get('security.context')->getToken()->getUser();
            //print_r($user->getRewardPoints());
            $point = '5';
            $rewrdPoint = $user->getRewardPoints() + $point;
            $user->setRewardPoints($rewrdPoint);
            $this->em->persist($user);
            $this->em->flush($user);
            //die;
        }
    }
    // Here you can intercept all HTTP requests, and through $container get access to user information
}
}

---

它与符号无关；最简单的方法是在会话中存储某个内容，并对其进行检查

比如，在打开登录用户的主页时，只需给他设置一个会话变量，并在处理请求侦听器之前检查会话变量

if($controller == $home && $this->container->get('session')->get('reward_was_payed', 0) != 1){
    $user = $this->container->get('security.context')->getToken()->getUser();
    //print_r($user->getRewardPoints());
    $point = '5';
    $rewrdPoint = $user->getRewardPoints() + $point;
    $user->setRewardPoints($rewrdPoint);
    $this->em->persist($user);
    $this->em->flush($user);

    $this->container->set('reward_was_payed', 1);
}

;

谢谢你的回答，你能再次看到我的问题吗？我变了$首页是数组。所以只要将if（$controller==$home&&$this->container->get（'session'）->get（'reward_was_payed'，0）！=1）{更改为if（in_array（$controller，$home）&&&$this->container->get（'session'）->get（'reward u was_was_payed'，0）！=1）{检查调试，如果设置了这个值，如果设置了session中的值，那么就不可能，get（'session get'）->get（'session（'reward_was_payed'，0）必须返回1；在本例中，它无法执行这部分代码