Question

在我的春季启动项目中，我有一个定义的聊天和用户实体，它们由多对多

相关联

Chat.java

@ManyToMany
@JoinTable(name = "chat_user",
        joinColumns = @JoinColumn(name = "chat_id", referencedColumnName = "id"),
        inverseJoinColumns = @JoinColumn(name = "user_id", referencedColumnName = "id"))
private Set<User> users = new HashSet<User>();

public Set<User> getUsers() {
    return users;
}

public void setUsers(Set<User> users) {
    this.users = users;
}

User.java

@ManyToMany
@JsonBackReference
@JoinTable(name = "chat_user",
        joinColumns = @JoinColumn(name = "user_id", referencedColumnName = "id"),
        inverseJoinColumns = @JoinColumn(name = "chat_id", referencedColumnName = "id"))
private Set<Chat> chats = new HashSet<Chat>();

public Set<Chat> getChats() {
    return chats;
}

public void setChats(Set<Chat> chats) {
    this.chats = chats;
}

如何在CrudRepository中编写方法，以便从用户列表中获得包含所有用户的聊天列表

到目前为止，我有一个书面方法可以返回一个聊天列表，其中包含用户列表中任意的用户

public interface ChatRepo extends CrudRepository<Chat, Long> {
List<Chat> findAllByUsers(User user);
List<Chat> findDistinctByUsersIn(Set<User> users);
}

Answer 1

没有直接的解决方案（从我看到的）。你能做的是什么像你这样在你的ChatRepo中创建一个方法

public interface ChatRepository extends CrudRepository<Chat, Long> {

  @Query("select c from Chat c WHERE :user in elements(c.users)")
  List<Chat> getChatsWithUsers(@Param("user") User user);
}

这将为您提供此特定用户所参与的所有聊天的结果。
然后，您可以为集合中的所有用户执行此操作，然后执行结果的交集为此，您需要将@ElementCollection放入聊天类

..........
@ElementCollection
  private Set<User> users = new HashSet<User>();

但我同意这不是一个理想的解决方案，因为如果列表太大，那么它会导致多个数据库调用

Answer 2

   @ManyToMany
 @JoinTable(name = "chat_user",
    joinColumns = @JoinColumn(name = "chat_id", referencedColumnName = "id"),
    inverseJoinColumns = @JoinColumn(name = "user_id", referencedColumnName = 
 "id"))
  private Set<User> users = new HashSet<User>();

  I think this is maybe wrong. Because one chat id cannot point out many user. 
  So you can use simple jointable to user entity

Answer 3

实现equals / hashCode就像Malte Hartwig建议的那样是解决方案

@Override
public boolean equals(Object o) {

    if (o == this) return true;
    if (!(o instanceof User)) {
        return false;
    }
    User user = (User) o;
    return id == user.id &&
            Objects.equals(firstName, user.firstName) &&
            Objects.equals(lastName, user.lastName) &&
            Objects.equals(userName, user.userName) &&
            Objects.equals(password, user.password);
}

@Override
public int hashCode() {
    return Objects.hash(id, firstName, lastName, userName, password);
}

如何找到该列表包含另一个列表中的所有元素？

3 个答案: