Question

I have a simple question regarding Entity declaration in JPA.
I have an entity with 2 foreign keys, which are not null and form an uniqueConstraint. First I was thinking about a composite key, composed of the two foreign keys, but I heard that this is a legacy design, and not the recommended way of designing new tables.

So I am interested if Hibernate/JPA can automatically generate id, based on the two foreign keys. Let’s say I have the following Entity:

@Entity
public class Foo {
  @ManyToOne
  private Bar bar;
  private int i;
}

(I omitted not null and uniqueConstraint tags to make the code more readable)

I know I can simply add an id field, with GeneratedValue, and let my DB generate the key (in my example MySQL with auto_increment), but this seems inefficient to me as it involves querying the database, and asking it to generate the unique id value.

Is there a way of generating an id, which is not composite (i.e. of type int or long), based on the id of the “Bar” class, and value of the integer “i”, since it those two values already form a unique constraint?

Answer 1

You may want to check out Chapter7 of “Java Persistence with Hibernate“.

You can model the composite key as an Embeddable:

import javax.persistence.*;
import java.io.Serializable;

@Entity
public class Foo {

    @Embeddable
    public static class Id implements Serializable {
        @Column(name = "bar_id_col")
        private Long barId;

        @Column(name = "i_col")
        private int i;

        public Id() {
        }

        public Id(Long barId, int i) {
            this.barId = barId;
            this.i = i;
        }

        @Override
        public boolean equals(final Object o) {
            if (this == o) {
                return true;
            }
            if (!(o instanceof Id)) {
                return false;
            }

            final Id id = (Id) o;

            if (i != id.i) {
                return false;
            }
            if (barId != null ? !barId.equals(id.barId) : id.barId != null) {
                return false;
            }

            return true;
        }

        @Override
        public int hashCode() {
            int result = barId != null ? barId.hashCode() : 0;
            result = 31 * result + i;
            return result;
        }
    }

    @EmbeddedId
    private Id id = new Id();

    @ManyToOne
    @JoinColumn(name = "bar_id_col", insertable = false, updatable = false)
    private Bar bar;

    private int i;

    public Foo() {
    }

    public Foo(Bar bar, int i) {
        // set fields
        this.Bar = bar;
        this.i=i;
        // set identifier values
        this.id.barId = bar.getId();
        this.id.i = i;
    }

}

Here I assume Bar looks like:

import javax.persistence.Entity;
import javax.persistence.Id;

@Entity
public class Bar {

    @Id
    Long id;

    public Long getId() {
        return id;
    }

    public void setId(final Long id) {
        this.id = id;
    }
}

Notice that this maps the bar_id_col twice. This is the reason for insertable = false, updatable = false in the second reference.

It’s tricky, but if you really want to do it like this, it’s possible.

Good luck,
J.