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.
+1
A:
I'm not sure this will fit your case, but give it a try.
@Entity
public class Friend {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private int friendId;
@Column
private String name;
@ManyToMany(mappedBy="owner")
private List<Friendship> friendships;
}
@Entity
public class Friendship {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private int friendshipId;
@OneToOne
private Friend owner;
@OneToOne
private Friend target;
// other info
}
Bozho
2009-12-02 10:12:19
+3
A:
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,
Arthur Ronald F D Garcia
2009-12-02 14:48:59
Thanks for your answer. But I have a question. Why is "insertable = false" put in the @ManyToOne relationships? If I remove it then it works. Otherwise the MyFriends is not persisted properly. I am not a Hibernate expert so probably I don't understand something.
Petar Minchev
2009-12-02 15:33:53
Ok got it, I haven't used an EmbeddedId like you but just another id column. You have put the insertable=false because the EmbeddedId takes care of it. Your answer is accepted:)
Petar Minchev
2009-12-02 15:41:01
You have noticed why. When two properties share the same column, it is a good idea to put settings about it in just one property. Otherwise, Hibernate will complain some errors.
Arthur Ronald F D Garcia
2009-12-02 15:50:04
A:
I had the same problem. You can try something like this:
<class name="Friend" entity-name="RelatedFriend" table="FRIENDS">
<id name="id" type="long">
<generator class="native" />
</id>
<!-- *** -->
<set name="relatedFriends" table="RELATED_FRIENDS">
<key column="FRIEND_ID" />
<many-to-many column="RELATED_FRIEND_ID" class="Friend" entity-name="RelatedFriend"/>
</set>
</class>
Bo
2010-05-02 09:56:47