java - JPA 多对多查询

Question

我有这样的类(class):

ChatRoom 类:

@Entity
public class ChatRoom {

    @Column(name="chatroom_id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long chatroomId;

    @ManyToMany(fetch=FetchType.EAGER)
    @JoinTable(
        name="CHATROOM_USER",
        joinColumns = {@JoinColumn(name="chatroom_id",referencedColumnName="chatroom_id")},
        inverseJoinColumns={@JoinColumn(name="user_id",referencedColumnName="user_id")}
    )
    private Set<User> users;

    @OneToMany(mappedBy="chatroom", cascade=CascadeType.PERSIST, orphanRemoval=true)
    private Set<ChatRoomMessage> messages;
}

用户类别:

@Entity
public class User {
    @Id
    @Column(name="user_id", length=40, unique=true, updatable=false)
    private String  userId;
}

ChatRoomMessage 类:

@Entity
public class ChatRoomMessage {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long messageId;

    @ManyToOne(optional=false)
    @JoinColumn(name="session_id", nullable=false)
    private User sender;

    @ManyToOne(optional=false)
    @JoinColumn(name="chatroom_id", nullable=true)
    private ChatRoom chatroom;
}

如果我想要用户发送的所有消息，我在 DAO 上做了这个:

MessagesDAO 类:

@Repository
public class MessageDaoImpl implements MessageDaoI {
    @Override
    public List<ChatRoomMessage> getMessages(User u){
        Criteria c = sessionFactory.getCurrentSession().createCriteria(ChatRoomMessage.class, "cmessage");
        c.createAlias("cmessage.sender", "sender");
        c.add(Restrictions.eq("sender", u));
        return c.list();
    }
}

当我尝试编写下一个查询时，问题就开始了:

1. 获取用户发送或接收的所有聊天室消息。
2. 获取用户在聊天室发送或接收的所有聊天室消息。

您能否提供有关如何解决疑问的想法？

Answer 1

您可以编写如下查询:

第一个查询:

Query query = session.createQuery("from ChatRoomMessage as c JOIN c.sender where c.sender.userId = :senderUserId");
query.setString("senderUserId", "SENDER_USER_ID_TO_BE_SEARCHED");
List list = query.list();

第二个查询:

Query query = session.createQuery("from ChatRoomMessage as c JOIN fetch c.chatroom where c.chatroom.users.userId = :senderUserId");
query.setString("senderUserId", "SENDER_USER_ID_TO_BE_SEARCHED");
List list = query.list();

希望这能解决您的问题。