Symfony 3获取实体内的当前用户

Question

我想知道是否有一种方法可以使用FOSUserBundle的实体User初始化属性所有者,以便它包含创建Post的用户

我想在构造函数中执行此操作,如下所示.

namespace AppBundle\Entity;

use Doctrine\ORM\Mapping as ORM;

/**
 * @ORM\Table(name="post")
 * @ORM\Entity(repositoryClass="AppBundle\Repository\PostRepository")
 */
class Post
{
  /* here are defined some attributs */
  /**
  * @ORM\ManyToOne(targetEntity="User", inversedBy="posts")
  * @ORM\JoinColumn(name="owner", referencedColumnName="id")
  */
  private $owner;

  public function __construct()
  {
    $this->owner = /* get current user */ ;
  }

}

有没有办法通过替换构造函数中的注释来做到这一点？

谢谢您的回答

Answer 1

不,没有.[*]

至少有两种方法可以解决这个问题:

• 通过填充owner属性的工厂服务创建Post实体 :

  namespace My\Bundle\EntityFactory;
  
  use Symfony\Component\Security\Core\Authentication\Token\Storage\TokenStorageInterface;
  use My\Bundle\Entity\Post;
  
  class PostFactory
  {
      private $tokenStorage;
  
      public function __construct(TokenStorageInterface $tokenStorage)
      {
          $this->tokenStorage = $tokenStorage;
      }
  
      public function createPost()
      {
          $user = $this->tokenStorage()->getToken()->getUser();
          $post = new Post($user);
      }
  }
  

  _{(对于此示例,您必须修改Post构造函数以接受所有者作为参数)}

  在services.yml中:

  services:
      post_factory:
          class: My\Bundle\EntityFactory\PostFactory
          arguments: [@security.token_storage]
  

  要从控制器创建实体:

  $post = $this->container->get('post_factory')->createPost();
  

• 如果您可以容忍只有在持久化实体后才设置所有者,则可以使用doctrine事件监听器:

  namespace My\Bundle\EventListener;
  
  use Doctrine\ORM\Event\LifecycleEventArgs;
  use Symfony\Component\Security\Core\Authentication\Token\Storage\TokenStorageInterface;
  use My\Bundle\Entity\Post;
  
  class PostOwnerAssignmentListener
  {
      private $tokenStorage;
  
      public function __construct(TokenStorageInterface $tokenStorage)
      {
          $this->tokenStorage = $tokenStorage;
      }
  
      public function prePersist(LifecycleEventArgs $event)
      {
          $entity = $event->getEntity();
          if ($entity instanceof Post && !$entity->getOwner()) {
              $entity->setOwner($this->tokenStorage->getToken()->getUser());
          }
      }
  }
  

  在services.yml中:

  services:
      post_owner_assignment_listener:
          class: My\Bundle\EventListener\PostOwnerAssignmentListener
          arguments: [@security.token_storage]
          tags:
              - { name: doctrine.event_listener, event: prePersit }
  

  这里的优点是无论Post创建方式和位置如何,都会分配所有者.

_{[*]:从技术上来说,默认情况下app.php你可以通过global $kernel;在你的构造函数中声明并从那里开始访问内核,然而这是非常强烈的劝阻并且可能以奇怪和微妙的方式打破.}

Answer 2

我认为你把这个问题过于复杂化了。当您在控制器中创建新帖子时，在控制器或存储库中执行如下操作：

use AppBundle\Entity\Post; //at top of controller

$em = $this->getDoctrine()->getManager();
$user = $this->container->get('security.token_storage')->getToken()->getUser();
$post = new Post();
$em->persist( $post );
$post->setOwner( $user );
// set other fields in your post entity
$em->flush();