检索用户的收藏夹 symfony

问题描述

我需要检索用户最近收藏的 9 个帖子。 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 Favorite look here 收藏夹

 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的控制器 ProfilController

/**
     * @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),]);
    }

这是出版物的经理PublicationManager

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，我很欣赏这些信息。谢谢

解决方法

我认为你不应该 pub.id = :pubId。

因此尝试使用第一个结果和最大结果并按顺序排序：

 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();
        
}