Symfony教條ManyToMany單向映射

Question

好吧，回頭再來，我會盡量簡化我的問題。

首先，我有2個實體

Post 
PostRating 

我已經創造了他們之間的單向多對多的關係，因爲我只需要評級被添加到每個崗位，如果我嘗試後映射到PostRating太，我收到循環引用錯誤。

郵政實體，它創建了3臺post_has_rating，PostRating實體內部沒有映射，它workes像預期，評級收集被添加到每個崗位，但如果我想找到一個等級，並進行必要的修改，然後它會比預期的更令人頭痛。

/** 
* Post have many PostRating 
* @ORM\ManyToMany(targetEntity="PostRating") 
* @ORM\JoinTable(name="post_has_rating", 
*  joinColumns={@ORM\JoinColumn(name="user_id", referencedColumnName="id")}, 
*  inverseJoinColumns={@ORM\JoinColumn(name="postrating_id", referencedColumnName="id", unique=true)} 
*  ) 
*/ 
protected $ratings; 

PostController中thumbAction，簡單的一句話 「ratingAction」

/** 
* Search related videos from youtube 
* @Route("/post/thumb", name="post_thumb") 
* @param Request $request 
* @return string 
*/ 
public function thumbAction (Request $request) { 
    $content = json_decode($request->getContent()); 
    $serializer = $this->get('serializer'); 
    $em = $this->getDoctrine()->getManager(); 

    $postRatingRepo = $this->getDoctrine()->getRepository(PostRating::class); 
    $postRepo = $this->getDoctrine()->getRepository(Post::class); 
    $me = $this->getUser()->getId(); 

    /** @var PostRating $rating */ 
    $rating = $postRatingRepo->findOneBy(['userId' => $me]); 

    /** @var Post $post */ 
    $post = $postRepo->find($content->id); 

    if ($post->getRatings()->contains($rating)) { 
     $post->removeRating($rating); 
     $em->remove($rating); 
    } 

    $rating = new PostRating(); 
    $rating->setUserId($me); 

    switch ($content->action) { 
    //NVM about those hardcoded words, they are about to be removed 
     case 'up': 
       $rating->setRating(1); 
      break; 
     case 'down': 
       $rating->setRating(0); 
      break; 
    } 

    $post->addRating($rating); 

     $em->persist($rating); 
     $em->persist($post); 
     $em->flush(); 

    return new JsonResponse($serializer->normalize(['success' => 'Post thumbs up created'])); 
} 

問題：$rating = $postRatingRepo->findOneBy(['userId' => $me]);該行需要有帖子ID也爲$post->getRatings()->contains($rating)，現在即時得到所有raitings ，我曾創建過，但它會拋出錯誤，如果我添加它，未知列

我應該創建自定義存儲庫，所以我可以使用DQL創建類似「findRating」的東西？

OR

我可以讓郵政和PostRating實體相互映射更簡單的方法，我真的不希望多到一對多的關係，因爲我沒有看到使用它

Answer 1

點

考慮你想保持一對多的單向這裏是我的建議

用於安置自己的實體

namespace AppBundle\Repository; 

use Doctrine\ORM\EntityRepository; 

class PostRepository extends EntityRepository 
{ 
    public function findOneRatingByUser($post, $user) 
    { 
     $query = $this->createQueryBuilder('p') 
      ->select('r') 
      ->innerJoin('p.ratings', 'r') 
      ->where('p.id = :post') 
      ->andWhere('r.user = :user') 
      ->setParameter('post', $post) 
      ->setParameter('user', $user) 
      ->getQuery() 
     ; 

     return $query->getOneOrNullResult(); 
    } 
} 

創建自定義庫然後喲烏爾控制器：

public function thumbAction (Request $request) 
{ 
    $content = json_decode($request->getContent()); 
    $serializer = $this->get('serializer'); 
    $em = $this->getDoctrine()->getManager(); 

    $postRepo = $this->getDoctrine()->getRepository(Post::class); 
    $me = $this->getUser()->getId(); 

    /** @var Post $post */ 
    $post = $postRepo->find($content->id); 

    $rating = $postRepo->findOneRatingByUser($post->getId(), $me); 

    if (null === $rating) { 
     $rating = new PostRating(); 
     $rating->setUserId($me); 
    } 

    switch ($content->action) { 
    //NVM about those hardcoded words, they are about to be removed 
     case 'up': 
       $rating->setRating(1); 
      break; 
     case 'down': 
       $rating->setRating(0); 
      break; 
    } 

    $post->addRating($rating); 

     $em->persist($rating); 
     $em->persist($post); 
     $em->flush(); 

    return new JsonResponse($serializer->normalize(['success' => 'Post thumbs up created'])); 
} 

如果您希望自定義庫的工作不要忘記聲明它在你的實體

/** 
* @ORM\Entity(repositoryClass="AppBundle\Repository\PostRepository") 
*/ 
class Post