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symfony - 检索用户的收藏夹 symfony

问题描述

我需要检索用户最后 9 个最喜欢的帖子。我有一个用户个人资料页面，其中列出了所有出版物以及每个用户创建的所有食谱，我有一个侧边栏，我想在其中显示最后 9 个收藏 夹 收藏夹看这里 FavoriteRepository

 public function lastXFavorite(Publication $publication, User $user, $nombre)
    {
        return $this->createQueryBuilder('f')
            ->join('f.publication', 'pub')
            ->join('f.user', 'user')
            ->andWhere('pub.id = :pubId AND user.id = :userId')
            ->setParameter('pubId', $publication->getId())
            ->setParameter('userId', $user->getId())
            ->setMaxResults($nombre)
            ->getQuery()
            ->getResult();
            
    }

这是我的用户个人资料页面 ProfilController的控制器

/**
     * @IsGranted("ROLE_USER")
     * @Route("/profil/{id}", name="profil", methods={"GET","POST"})
     * 
     */
    public function index(User $user, PublicationManager $publicationManager): Response
    {
        // historique des les publications
        $em = $this->getDoctrine()->getManager();
        $publications = $em->getRepository('App:Publication')->findBy(
            array('users' => $user->getId()),
            array('created_at' => 'Desc')
        );

        // Lister all mes Favorites
        $favorites = $em->getRepository('App:Favorite')->findBy(
            array('user' => $user->getId())
           
        ); 

        // last 9 favorites
        
        
        

        return $this->render('profil/index.html.twig', [

            'publications' => $publications,
            'mesRecettes' => $publicationManager->getRecetteByUser($user),
            'mesFavoris' => $favorites,
            'recettes' => $publicationManager->allRecette(),
            'user' => $user,
            'lastRecettes' => $publicationManager->lastXRecette(),
            'lastPRecettes' => $publicationManager->lastPRecette($user, 6),
               

        ]);
    }

这是出版物PublicationManager的经理

class PublicationManager
{
    protected $em;

    public function __construct(EntityManagerInterface $entityManager)
    {
        $this->em = $entityManager;
    }

    public function getPublicationByUser(User $user)
    {
        return $this->em->getRepository(Publication::class)->getPublicationUser($user);
    }

    public function getRecetteByUser(User $user)
    {
        return $this->em->getRepository(Publication::class)->findBy(
            ['type' => true, 'users' => $user->getId()],
            ['created_at' => 'desc']
        );
    }

    public function allRecette()
    {
        return $this->em->getRepository(Publication::class)->findBy(
            ['type' => true],
            ['created_at' => 'desc']
        );
    }

    public function allPublication()
    {
        return $this->em->getRepository(Publication::class)->findBy([], ['created_at' => 'desc']);
    }

    public function lastXRecette()
    {
        return $this->em->getRepository(Publication::class)->lastXRecette(4);
    }

    public function lastPRecette(User $user, $cant)
    {
        return $this->em->getRepository(Publication::class)->lastPRecette($user, $cant);
    }

我是初学者，我很感激这些信息。谢谢

标签： symfonysymfony4symfony5

解决方案

我认为你不应该在哪里pub.id = :pubId。

因此，尝试使用第一个结果和最大结果并按以下顺序排序：

 public function lastXFavorite(Publication $publication, User $user, $nombre)
{
    return $this->createQueryBuilder('f')
        ->join('f.publication', 'pub')
        ->join('f.user', 'user')
        ->andWhere('user.id = :userId')
        ->setParameter('userId', $user->getId())
        ->orderBy('f.id', 'DESC')
        ->setFirstResult(0)
        ->setMaxResults(9)
        ->getQuery()
        ->getResult();
        
}

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