在同一张桌子上多对多地添加列 [英] Many-to-many on the same table with additional columns

问题描述

我有一个班级用户。用户可以成为许多其他用户的朋友。这种关系是相互的。如果A是B的朋友，那么B是A的朋友。另外，我希望每个关系都存储其他数据 - 例如两个用户成为朋友的日期。所以这是在同一张桌子上的多对多关系，并增加了一些列。我知道应该创建一个中产阶级友谊（包含两个用户ID和日期栏）。但是我将这个映射到Hibernate中来。阻止我的是映射到同一个表。我可以解决它，如果多对多关系是在两个不同的表之间。

多对多关系在同一张桌子上

这不是一个好主意。这是一个噩梦维持。

尝试这一个，而不是

  @Entity 
 public class Friend {
 
 @Id 
 @GeneratedValue（strategy = GenerationType.AUTO）
 private Integer friendId; 
 
 @Column 
私人字符串名称; 
 
 @OneToMany（mappedBy =我）
私人列表< MyFriends>我的朋友; 
 
} 
 
 @Entity 
 public class MyFriends {
 
 @EmbeddedId 
 MyFriendsId id; 
 
 @Column 
 private String additionalColumn; 
 
 @ManyToOne 
 @JoinColumn（name =ME_ID，insertable = false，updateable = false）
 private Friend me; 
 
 @ManyToOne 
 @JoinColumn（name =MY_FRIEND_ID，可插入= false，可更新= false）
私人朋友myFriend; 
 
 @Embeddable 
 public static class MyFriendsId implements Serializable {
 
 @Column（name =ME_ID，nullable = false，updateable = false）
私人整数meId; 
 
 @Column（name =MY_FRIEND_ID，nullable = false，updateable = false）
 private Integer myFriendId; 
 $ b $ public boolean equals（Object o）{
 if（o == null）
 return false; 
 
 if（！（o instanceof MyFriendsId））
 return false; 
 
 MyFriendsId other =（MyFriendsId）o;如果（！（other.getMeId（）。equals（getMeId（）））
返回false; 
 
 if（！（other.getMyFriendId（）。equals（getMyFriendId ）））
返回false; 
 
返回true; 
} 
 
 public int hashcode（）{
 // hashcode impl 
 
 
 
 
 $ b} 
  

问候，

I have a class User. A user can be a friend with many other users. The relationship is mutual. If A is a friend of B then B is a friend of A. Also I want every relation to store additional data - for example the date when two users became friends. So this is a many-to-many relationship on the same table with additional columns. I know that a middle class Friendship should be created(containing two user ids and column for the date). But I am coming short at mapping this with Hibernate. The thing that stops me is that the mapping is to the same table. I can solve it, if the many-to-many relationship was between two different tables.

解决方案

You have said

many-to-many relationship on the same table

It is not a good idea. It is a nightmare to maintain.

Try this one instead

@Entity
public class Friend {

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    private Integer friendId;

    @Column
    private String name;

    @OneToMany(mappedBy="me")
    private List<MyFriends> myFriends;

}

@Entity
public class MyFriends {

    @EmbeddedId
    private MyFriendsId id;

    @Column
    private String additionalColumn;

    @ManyToOne
    @JoinColumn(name="ME_ID", insertable=false, updateable=false)
    private Friend me;

    @ManyToOne
    @JoinColumn(name="MY_FRIEND_ID", insertable=false, updateable=false)
    private Friend myFriend;

    @Embeddable
    public static class MyFriendsId implements Serializable {

        @Column(name="ME_ID", nullable=false, updateable=false)
        private Integer meId;

        @Column(name="MY_FRIEND_ID", nullable=false, updateable=false)
        private Integer myFriendId;

        public boolean equals(Object o) {
            if(o == null)
                return false;

            if(!(o instanceof MyFriendsId))
                return false;

            MyFriendsId other = (MyFriendsId) o;
            if(!(other.getMeId().equals(getMeId()))
                return false;

            if(!(other.getMyFriendId().equals(getMyFriendId()))
                return false;

            return true;
        }

        public int hashcode() {
            // hashcode impl
        }

    }


}

regards,

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