如果孩子为空，休眠ManyToMany删除另一边

Posted 2023-02-24

【中文标题】如果孩子为空，休眠ManyToMany删除另一边

【英文标题】：Hibernate ManyToMany remove other side if children empty

【发布时间】：2021-10-24 04:03:26

【问题描述】：

我有 2 个具有 @ManyToMany 关系的实体。如果users 字段为空，我想在删除User 的同时删除Activity。

@Entity
public class User 
...
@ManyToMany(mappedBy="users")
private List<Activity> activities;

和

@Entity
public class Activity 

@ManyToMany(cascade = CascadeType.PERSIST, CascadeType.MERGE, CascadeType.REFRESH, CascadeType.DETACH)
@JoinTable(name = "USER_ACTIVITY",
           joinColumns = @JoinColumn(name = "ACTIVITY_ID"),
           inverseJoinColumns = @JoinColumn(name = "USER_ID"))
private List<User> users;

例如删除前有以下记录

USERS
id    name
----+-------
1     user1
2     user2
3     user3

ACTIVITY
id    name
----+-------
1     act1
2     act2
3     act3

USER_ACTIVITY
USER_ID    ACTIVITY_ID
---------+-------
1          1
2          1
2          2
3          1
3          3
2          3
1          3

如果user2 被删除，那么act2 也应该被删除。因为它没有更多的用户。所以删除 user2 后的记录应该如下所示：

USERS
id    name
----+-------
1     user1
3     user3

ACTIVITY
id    name
----+-------
1     act1
3     act3

USER_ACTIVITY
USER_ID    ACTIVITY_ID
---------+-------
1          1
3          1
3          3
1          3

【参考方案1】：

您必须反转映射并像这样使用CascadeType.ALL：

@Entity
public class User 
...
  @ManyToMany(cascade = CascadeType.ALL)
  @JoinTable(name = "USER_ACTIVITY",
           joinColumns = @JoinColumn(name = "USER_ID"),
           inverseJoinColumns = @JoinColumn(name = "ACTIVITY_ID"))
  private List<Activity> activities;


@Entity
public class Activity 

  @ManyToMany(mappedBy="activities")
  private List<User> users;