JPA OneToMany and ManyToOne throw: Repeated column in mapping for entity column (should be mapped with insert="false" update="false")

Question

I have three classes one of the names is User and this user has other classes instances. Like this;

public class User{
    @OneToMany(fetch=FetchType.LAZY, cascade = CascadeType.ALL)
    public List<APost> aPosts;

    @OneToMany(fetch=FetchType.LAZY, cascade = CascadeType.ALL)
    public List<BPost> bPosts;
}




   public class BPost extends Post {
    
    @ManyToOne(fetch=FetchType.LAZY)    
    public User user;
 }
    
    public class APost extends Post {
    
     @ManyToOne(fetch=FetchType.LAZY) 
     public User user;
 }

it's working like this but generates empty tables in DB. Which have to contain foreign keys. When I tried to use mappedBy and JoinColumn annotations I got failed. How can I resolve this?

Extra information:

When I changed with;

 @ManyToOne(fetch=FetchType.LAZY)
 @JoinColumn(name="id")
 public User user;

and

 @OneToMany(fetch=FetchType.LAZY, cascade = CascadeType.ALL, mappedBy="id")
 public List<APost> aPosts;

I'm getting

A JPA error occurred (Unable to build EntityManagerFactory): Repeated column in mapping for entity: models.post.APost column: id (should be mapped with insert="false" update="false")

Final Edit: Finally, I was totally wrong about JPA annotations. :( When I change

@OneToMany(fetch=FetchType.LAZY, cascade = CascadeType.ALL, mappedBy="id")

to

@OneToMany(fetch=FetchType.LAZY, cascade = CascadeType.ALL, mappedBy="user")

and

@ManyToOne(fetch=FetchType.LAZY)
@JoinColumn(name="user_id")

everything works ok. :)

Answer 1

I am not really sure about your question (the meaning of "empty table" etc, or how mappedBy and JoinColumn were not working).

I think you were trying to do a bi-directional relationships.

First, you need to decide which side "owns" the relationship. Hibernate is going to setup the relationship base on that side. For example, assume I make the Post side own the relationship (I am simplifying your example, just to keep things in point), the mapping will look like:

(Wish the syntax is correct. I am writing them just by memory. However the idea should be fine)

public class User{
    @OneToMany(fetch=FetchType.LAZY, cascade = CascadeType.ALL, mappedBy="user")
    private List<Post> posts;
}


public class Post {
    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="user_id")
    private User user;
}

By doing so, the table for Post will have a column user_id which store the relationship. Hibernate is getting the relationship by the user in Post (Instead of posts in User. You will notice the difference if you have Post's user but missing User's posts).

You have mentioned mappedBy and JoinColumn is not working. However, I believe this is in fact the correct way. Please tell if this approach is not working for you, and give us a bit more info on the problem. I believe the problem is due to something else.

---

Edit:

Just a bit extra information on the use of mappedBy as it is usually confusing at first. In mappedBy, we put the "property name" in the opposite side of the bidirectional relationship, not table column name.

Answer 2

You should never use the unidirectional @OneToMany annotation because:

1. It generates inefficient SQL statements
2. It creates an extra table which increases the memory footprint of your DB indexes

Now, in your first example, both sides are owning the association, and this is bad.

While the @JoinColumn would let the @OneToMany side in charge of the association, it's definitely not the best choice. Therefore, always use the mappedBy attribute on the @OneToMany side.

public class User{
    @OneToMany(fetch=FetchType.LAZY, cascade = CascadeType.ALL, mappedBy="user")
    public List<APost> aPosts;

    @OneToMany(fetch=FetchType.LAZY, cascade = CascadeType.ALL, mappedBy="user")
    public List<BPost> bPosts;
}

public class BPost extends Post {

    @ManyToOne(fetch=FetchType.LAZY)    
    public User user;
}

public class APost extends Post {

     @ManyToOne(fetch=FetchType.LAZY) 
     public User user;
}