嵌入式id成员成员值的Hibernate标准

Question

I would like to find an entity using a critera with restriction on the value of an attribute of a second entity wich is a member of the embedded id of my first entity.

First entity :

@Entity
public class Car {

    @EmbeddedId
    private Id id = new Id();

    private String color;

    @Embeddable
    public static class Id implements Serializable {

        private static final long serialVersionUID = -8141132005371636607L;

        @ManyToOne
        private Owner owner;

        private String model;

        // getters and setters...
        // equals and hashcode methods

    }

    // getters and setters...

}

Second entity :

@Entity
public class Owner {

    @Id
    @GeneratedValue (strategy = GenerationType.AUTO)
    private Long id;
    private String firstname;    
    private String lastname;

    @OneToMany (mappedBy = "id.owner")
    private List<Car> cars;

    // getters and setters...

}

In this example, I would like to obtain the car with the color 'black', model 'batmobile' and the owner's firstname 'Bruce' (oops... spoiler ;) )

I tried to do something like that but it won't work :

List<Car> cars = session.createCriteria(Car.class)
      .add(Restrictions.eq("color", "black"))
      .add(Restrictions.eq("id.model", "batmobile"))
      .createAlias("id.owner", "o")
      .add(Restrictions.eq("o.firstname", "Bruce"))
      .list();

Result :

Hibernate: select this_.model as model1_0_0_, this_.owner_id as owner_id3_0_0_, this_.color as color2_0_0_ from Car this_ where this_.color=? and this_.model=? and o1_.firstname=?
ERROR: Unknown column 'o1_.firstname' in 'where clause'

What is the right way to obtain what I want ?

update

I tried in hql :

String hql = "FROM Car as car where car.color = :color and car.id.model = :model and car.id.owner.firstname = :firstname";

Query query = em.createQuery(hql);
query.setParameter("color", "black");
query.setParameter("model", "batmobile");
query.setParameter("firstname", "Bruce");

List<Car> cars = query.getResultList();

It works but is there a way to do this with criteria ?

Answer 1

You forgot to add the @Column annotation on top of the firstname and lastname fields (and the color field in Car). In hibernate if a field is not annotated, it doesn't recognize it as a database field. This page should give you a good idea about how to set up your model objects .

NOTE: You can have the column annotation over the getters and be fine, but you didn't show the getters. Either place is fine.

---

Look at what HQL is spitting back out, specifically the statement (formated for easier reading):

select 
    this_.model as model1_0_0_, 
    this_.owner_id as owner_id3_0_0_, 
    this_.color as color2_0_0_ 
from Car this_ 
where 
    this_.color=? 
    and this_.model=? 
    and o1_.firstname=?

It looks like hibernate is translating the field "id.owner" to "o" as your alias told it to to, but for some reason it's not writing down that "id.owner=o" as intended. You may want to do some research into why it may be doing that.

Answer 2

As per https://hibernate.atlassian.net/browse/HHH-4591 there is a workaround.

You have to copy the needed relation-property of the @EmbeddedId ( owner in this case) to the main entity ( Car in this case) with insertable = false, updatable = false as follows

@Entity
public class Car {

    @EmbeddedId
    private Id id = new Id();

    private String color;

    @ManyToOne
    @JoinColumn(name = "column_name", insertable = false, updatable = false)
    private Owner owner;

    @Embeddable
    public static class Id implements Serializable {

        private static final long serialVersionUID = -8141132005371636607L;

        @ManyToOne
        private Owner owner;

        private String model;

        // getters and setters...
        // equals and hashcode methods

    }

    // getters and setters...

}

Then just create directly the alias instead of using the composite id property

List<Car> cars = session.createCriteria(Car.class)
      .add(Restrictions.eq("color", "black"))
      .add(Restrictions.eq("id.model", "batmobile"))

      .createAlias("owner", "o")

      .add(Restrictions.eq("o.firstname", "Bruce"))
      .list();