0

我有这门课：

    @Entity
    @Table(name = "USERS")
    public class User {

    @Id
    @Column(name = "USER_ID")
    @GeneratedValue
    private long userId;
    ...
    @ManyToMany(fetch = FetchType.LAZY)
    @JoinTable(name = "FRIENDS", joinColumns = @JoinColumn(name = "USER_ID"), inverseJoinColumns = @JoinColumn(name = "FRIEND_ID"))
        private Set<User> friends;


    @ManyToMany(fetch = FetchType.LAZY)
    @JoinTable(name = "FRIENDS", joinColumns = @JoinColumn(name = "FRIEND_ID"), inverseJoinColumns = @JoinColumn(name = "USER_ID"))
        private Set<User> friendOf;

   ...
    public Set<User> getAllFriends() {
      allFriends = new HashSet<User>();
      allFriends.addAll(friends);
      allFriends.addAll(friendOf);
      return allFriends;
}
   // getters and seters

如果我需要用户的所有朋友，我可以通过简单地调用 getAllFriends() 来获得它们。但是我想对返回朋友的最大数量添加一些限制。所以我想选择所有使用 HQL 的朋友。我想要这样的东西（我的 hql 不正确，只是为了展示这个想法）：

String hql = "Select u FROM User u inner join u.friendOf fof WHERE fof.userId = :userId inner join u.friends fs WHERE fs.userId = :userId";
    Query query = sessionFactory.getCurrentSession().createQuery(hql);
    query.setParameter("userId", userId);
    List<User> results = query.setMaxResults(maxResults).setFirstResult(firstResult).list();

是否可以创建这样的查询？

更新：我有符合我要求的 SQL 查询。也许有人可以帮我把它翻译成 HQL？

select * from users where user_id in (
select 
 (case  
  when user_id = :userId then friend_id 
  else user_id 
 end) as id
from friends
where user_id = :userId or friend_id = :userId);
4

1 回答 1

2

HQL 不支持UNION子句（存在未解决的问题：https ://hibernate.atlassian.net/browse/HHH-1050 ），所以这有点难，但我认为你可以这样做：

select u
from User u
where u.userId in (
        select fr.userId
        from User u1
        inner join u1.friends fr
        where u1.userId = :userId
    ) 
    or u in (
        select fr.userId
        from User u2
        inner join u2.friendOf fr
        where u2.userId = :userId
    )
于 2013-06-13T02:15:22.673 回答