CODEWITHSUNDEEP
javahibernatejpa
ramiromd
ramiromd

Reputation: 2029

Querying by inherited property

i am trying to do my first steps in Java, and i am blocked with the next problem: i have this hieararchy class "User -> Traveller".

@Entity
@Inheritance(strategy=InheritanceType.JOINED)
public class User {
   // Some properties and accessors.
}

@Entity
@PrimaryKeyJoinColumn(name="user_id")
public class Traveller extends User{ 
   // Some properties and accessors.
}

So, i need search a traveller by your username which is a User class property. I try this in my TravellerDao:

public Traveller findByUsername(String username){

    EntityManager em = PersistenceHelper.getEm();
    TypedQuery<Traveller> query = em.createQuery("FROM Traveller WHERE username = ?1", Traveller.class);
    query.setParameter(1, username);

    return query.getSingleResult();
}

But, a javax.persitence.NoResultException is catched. Where is the problem ? In the "where" clausule ? In the join ?. Any ideas ?

Notes:

1) The hierarchy works fine with "find" method. And the database is generated correctly.

2) I have an entry in the Traveller table that links up to one in User table.

3) In the database exists an user with the username that i am using to test.

4) The method get correctly the "username" parameter.

EDIT: Now works. The problem was the execution order of test cases. The query works fine.

Upvotes: 0

Views: 156

Answers (1)

Jazzepi
Jazzepi

Reputation: 5480

I think your query is incorrect, I believe you are forced to name an alias for every entity you want to interact with. Have you tried this?

"SELECT t FROM Traveller t WHERE t.username = ?1", Traveller.class

Also you should have to cast the result when you call query.getSingleResult(). Because you pass in Traveller.class into the createQuery call which is paramaterized by Class<T> resultClass

Edit: Added a 1 after the ? as shown here. http://www.objectdb.com/java/jpa/query/parameter#Ordinal_Parameters_index_

 public Country getCountryByName(EntityManager em, String name) {
    TypedQuery<Country> query = em.createQuery(
        "SELECT c FROM Country c WHERE c.name = ?1", Country.class);
    return query.setParameter(1, name).getSingleResult();
  }

Upvotes: 1

Related Questions