Question

我有一个用户模型，我希望包含一个＆＃34;朋友列表＆＃34;：

@Entity 
public class User extends Model {

  @Id
  public long id; 

  @ManyToMany
  public List<User> friends;

  ...
}

运行它会给我一个错误（Database 'default' is in an inconsistent state）和一个包含下表的进化文件：

create table user_user (
    user_id                        bigint not null,
    user_id                        bigint not null,
    constraint pk_user_user primary key (user_id, user_id)
)

我该如何解决这个问题？另外，对于奖励积分，我如何获得与我的用户成为朋友的人员列表？

Answer 1

我不建议将ManyToMany用于实际项目中的self实体。最好创建另一个表，比如“友谊”。看看这个问题的答案：Many-to-many on the same table with additional columns

任何方式，这里都是你问题的答案：

您需要“通过武器”配置连接表

package models;

import java.util.*;
import javax.persistence.*;

import com.avaje.ebean.Model;

@Entity
public class User extends Model {

  @Id
  public long id;

  @ManyToMany(targetEntity=User.class)
  @JoinTable(name="friends",
      joinColumns={@JoinColumn(name="user_a_id", referencedColumnName="id")},
      inverseJoinColumns={@JoinColumn(name="user_b_id", referencedColumnName="id")}
  )
  public List<User> friends;
}

自动生成的进化脚本：

create table user (
  id                        bigint not null,
  constraint pk_user primary key (id))
;


create table friends (
  user_a_id                      bigint not null,
  user_b_id                      bigint not null,
  constraint pk_friends primary key (user_a_id, user_b_id))
;
create sequence user_seq;




alter table friends add constraint fk_friends_user_01 foreign key (user_a_id) references user (id) on delete restrict on update restrict;

alter table friends add constraint fk_friends_user_02 foreign key (user_b_id) references user (id) on delete restrict on update restrict;

# --- !Downs

SET REFERENTIAL_INTEGRITY FALSE;

drop table if exists user;

drop table if exists friends;

SET REFERENTIAL_INTEGRITY TRUE;

drop sequence if exists user_seq;

如何定义同一模型的ManyToMany关系？

1 个答案: