计算多对多表休眠中ID的出现

Question

我正在创建一个函数，其中我的应用程序将显示最喜欢的照片。

我有这堂课

public class User{
    @OneToMany(fetch = FetchType.LAZY, mappedBy = "pk.user", cascade = CascadeType.ALL)
    private Set<UserLikedPhoto> likedPhotos = new HashSet<UserLikedPhoto>();
   //Other properties and accessor methods.
}

摄影班

public class Photo{
    @OneToMany(fetch = FetchType .LAZY ,mappedBy = "pk.photo")
    private Set<UserLikedTrack> likedByUsers = new HashSet<UserLikedTrack>();
        //Other properties and accessor methods
}

CompoundId / CompoundObject

@Embeddable
public class UserLikedPhotoId  implements Serializable {

    @ManyToOne
    private UserProfile user;

    @ManyToOne
    private Photo photo;
    //Other Properties and accessor methods.
}

以及包含CompoundObject和日期的类

  @Entity
    @AssociationOverrides({
            @AssociationOverride(name = "pk.userId", joinColumns = @JoinColumn(name = "userId")),
            @AssociationOverride(name = "pk.photoid", joinColumns = @JoinColumn(name = "photoId")) })
    public class UserLikedPhoto{

        @EmbeddedId
        private UserLikedPhotoId pk = new UserLikedPhotoId();
        @Column
        @Temporal(TemporalType.DATE)
        private Date date;

           //Other Entities and accssor methods
    }

有了这个班。 我会用这种类型的表生成

------------------------------
| date |    UserId   photoId |
-----------------------------
| 2010-12-23  | 1   | 23 |
| 2010-12-21  | 2   | 23 |
| 2010-12-23  | 1   | 24 |
| 2010-12-21  | 5   | 23 |

现在，我想做的就是在示例中获得投票最多的照片（在给定日期前5名或前10名），投票最多的照片是23号照片，投票第二多的照片是24号照片。

在Hibernate中，我将如何查询此类任务？

Answer 1

哈芬特里这..但只是尝试...如果没有用，我会删除这个答案

select photoId, count(photoId) from UserLikedPhoto group by photoId order by count(photoid) desc

在此查询中，我们按photoid进行分组，因此1张照片id只会有1行。

UserLikedPhoto提供的这张照片ID会告诉我们该照片被点UserLikedPhoto次数的计数...

我们正在按desc订购，因此最喜欢的将排在最前面。 您问您要先5还是10 ...所以您可以在select语句中使用前10或前5个sql子句

所以最终的SQL将是这样的...

select top 10 photoId, count(photoId) from UserLikedPhoto group by photoId order by count(photoid) desc

Hibernate确实也支持本机sql。

Answer 2

如果没有测试，从一开始就很难做到，但是这是我的尝试：

Query q = session.createQuery("select p from 
    Photo as p, 
    (select ulp.pk.photo.photoId as photoId from UserLikedPhoto ulp 
           where ulp.date = :someDate group by ulp.pk.photo.photoId 
                 order by count(ulp.pk.photo.photoId) desc) as top 
    where p.photoId = top.photoId");
q.setParameter("someDate", new Date())
q.setMaxResults(10);
List<Photo> topPhotos = q.list();