JPA - Criteria API and EmbeddedId

Question

I want to use criteria to make the following query. I have an Entity with EmbeddedId defined:

 @Entity  @Table(name="TB_INTERFASES")  public class Interfase implements Serializable {    @EmbeddedId   private InterfaseId id;  }   @Embeddable  public class InterfaseId implements Serializable {   @Column(name="CLASE")   private String clase;  } 

And the criteria query that i am trying to do is:

 CriteriaBuilder criteriaBuilder = this.entityManager.getCriteriaBuilder();  CriteriaQuery<Interfase> criteriaQuery = criteriaBuilder.createQuery(Interfase.class);  Root<Interfase> entity = criteriaQuery.from(Interfase.class);  criteriaQuery.where(    criteriaBuilder.equal(entity.get("clase"), "Clase"),  ); 

But this is throwing an IllegalArgumentException:

java.lang.IllegalArgumentException: Not an managed type: class InterfaseId 

i've tried with this queries too:

 Root<Interfase> entity = criteriaQuery.from(Interfase.class);  criteriaQuery.where(    criteriaBuilder.equal(entity.get("id").get("clase"), "Clase"),  ); 

and this one too...

 Root<Interfase> entity = criteriaQuery.from(Interfase.class);  criteriaQuery.where(    criteriaBuilder.equal(entity.get("id.clase", "Clase"),  ); 

with no luck. So my question is how can i make a query with criteria when my classes are using Embedded and EmbeddedId annotations?

Thanks!. Mauro.

Answer 1

You need to use path navigation to access the attribute(s) of the Embeddable. Here is an example from the JPA 2.0 specification (using the static metamodel):

6.5.5 Path Navigation

...

In the following example, ContactInfo is an embeddable class consisting of an address and set of phones. Phone is an entity.

CriteriaQuery<Vendor> q = cb.createQuery(Vendor.class); Root<Employee> emp = q.from(Employee.class); Join<ContactInfo, Phone> phone =     emp.join(Employee_.contactInfo).join(ContactInfo_.phones); q.where(cb.equal(emp.get(Employee_.contactInfo)                     .get(ContactInfo_.address)                     .get(Address_.zipcode), "95054"))     .select(phone.get(Phone_.vendor)); 

The following Java Persistence query language query is equivalent:

SELECT p.vendor FROM Employee e JOIN e.contactInfo.phones p WHERE e.contactInfo.address.zipcode = '95054' 

So in your case, I think you'll need something like this:

criteriaBuilder.equal(entity.get("id").get("clase"), "Referencia 111") 

References

• JPA 2.0 Specification
  • Section 6.5.5 "Path Navigation"

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Update: I've tested the provided entities with Hibernate EntityManager 3.5.6 and the following query:

CriteriaBuilder builder = em.getCriteriaBuilder();  CriteriaQuery<Interfase> criteria = builder.createQuery(Interfase.class); Root<Interfase> interfaseRoot = criteria.from(Interfase.class); criteria.select(interfaseRoot); criteria.where(builder.equal(interfaseRoot.get("id").get("clase"),      "Referencia 111"));  List<Interfase> interfases = em.createQuery(criteria).getResultList(); 

runs fine and generates the following SQL:

 17:20:26.893 [main] DEBUG org.hibernate.SQL -      select         interfase0_.CLASE as CLASE31_      from         TB_INTERFASES interfase0_      where         interfase0_.CLASE=? 17:20:26.895 [main] TRACE org.hibernate.type.StringType - binding 'Referencia 111' to parameter: 1 

Works as expected.