Symfony2侦听器与DataFixture

Question

我已将以下侦听器注册为服务。这将保存已登录的用户。它工作得很好。保存实体后，用户id位于createdBy和updatedBy中。好的，有一个小问题：命令“php-app/console-doctrine:fixtures:load”对非对象上的成员函数getUser（）抛出错误“调用。这是可以理解的。直到现在我才必须在以前每次禁用该服务？你有其他解决方案吗？

class UserListener implements EventSubscriber
{
    protected $container;

    public function __construct(ContainerInterface  $container)
    {
        $this->container = $container;
    }

    public function getSubscribedEvents()
    {
        return array(
            Events::prePersist,
            Events::preUpdate
        );
    }

    public function prePersist(LifecycleEventArgs $args)
    {
        $entity = $args->getEntity();

        if ($entity instanceof Post) {

            $user = $this->container->get('security.context')->getToken()->getUser();

            if (!is_object($user) || !$user instanceof User) {
                throw new AccessDeniedException();
            }

            $entity->setCreatedBy($user);
            $entity->setUpdatedBy($user);
        }
    }

    /**
     * @param PreUpdateEventArgs $args
     */
    public function preUpdate(PreUpdateEventArgs $args)
    {
        $entity = $args->getEntity();

        $em = $args->getEntityManager();

        if ($entity instanceof Post) {

            $user = $this->container->get('security.context')->getToken()->getUser();

            if (!is_object($user) || !$user instanceof User) {
                throw new AccessDeniedException();
            }

            $entity->setUpdatedBy($user);

            $uow = $em->getUnitOfWork();
            $meta = $em->getClassMetadata(get_class($entity));
            $uow->recomputeSingleEntityChangeSet($meta, $entity);
        }
    }
}

Answer 1

我的最终代码（Symfony 2.6）：

1）听者：

<?php

namespace AppBundle\Listener;

use AppBundle\Entity\Post;
use AppBundle\Entity\Comment;

use Doctrine\ORM\Event\LifecycleEventArgs;
use Symfony\Component\Security\Core\Authentication\Token\Storage\TokenStorageInterface;

/**
 * Class BlameableListener
 *
 * @package AppBundle\Listener
 */
class BlameableListener
{
    private $tokenStorage;

    public function __construct(TokenStorageInterface $tokenStorage)
    {
        $this->tokenStorage = $tokenStorage;
    }

    public function prePersist(LifeCycleEventArgs $args)
    {
        $entity = $args->getEntity();

        if ($entity instanceof Post || $entity instanceof Comment) {
            // is authentication information available?
            if (null !== $this->tokenStorage->getToken()) {
                // get User
                $user = $this->tokenStorage->getToken()->getUser();
                if (is_object($user)) {
                    $entity->setAuthorEmail($user->getEmail());
                }
            }
        }
    }
}

2） 配置为服务

services:
    app.blameable.listener:
        class: AppBundle\Listener\BlameableListener
        arguments:
            - "@security.token_storage"
        tags:
            - { name: doctrine.event_listener, event: prePersist }

Answer 2

我有一个与类似的侦听器，并试图保存一个（模拟）实体（在您的示例中为“Post”）。当然，从控制台启动命令时我没有登录。

我给监听器添加了一个条件：

if ($entity instanceof Post) {
    if (null !== $entity->getUser() {
        return;
    }
}

现在，我可以在装置内的Post实体上设置（模拟）用户：

$user = $em->getReference('Application\Sonata\UserBundle\Entity\User', 1);
$post = new Post();
$post->setUser($user);

仍然不确定在实体侦听器中设置用户是否是好的做法。

我想目前大多数人会推荐使用ValueObject的"Post"或至少需要设置的构造函数，例如"name"和"User"。

Answer 3

调用“加载装置”命令时，您未登录。也许$this-