How to map an abstract class or interface in Hibernate HQL?

Imagine that I have a Debtor class. With Hibernate, I will define the class like that:

@Entity
@Table(name = "T_DEBTOR")
public class Debtor {

    @Id
    @Column(name = "ID_DEBTOR")
    private String idDebtor;
    ...

My DAO will then looks like:

public class DebtorDaoImpl implements DebtorDao {

    @PersistenceContext
    private EntityManager em;

    @SuppressWarnings("unchecked")
    public List<Debtor> findAllDebtors() {
        Query q = em.createQuery("select d from Debtor d");
        return (List<Debtor>) q.getResultList();
    }

This works well. However, I am in a configuration where I need to access differents schemas (as pointed here). Of course, in each schema the table that hosts the debtor list has not the same name. In addition to that, they may not have the exact same structure. That why I have x differents Debtor class (where x is the number of schemas I manipulate).

In the case where I have two differents schemas, I will have two different Debtor class: DebtorOne and DebtorTwo. As I want to ease my developments开发者_运维问答, I created an Interface (or an Abstract class, it does not change my problem here) that is implemented by both DebtorOne and DebtorTwo:

public interface Debtor {

    String getIdDebtor();

}

and:

@Entity
@Table(name = "T_DEBTOR_ONE")
public class DebtorOne implements Debtor {

    @Id
    @Column(name = "ID_DEBTOR")
    private String idDebtor;
    ...

If I let my DAO as it is, I get the following error from Hibernate:

Caused by: org.hibernate.hql.ast.QuerySyntaxException: Debtor is not mapped [select d from Debtor d]

If I change my DAO to have this:

    public List<Debtor> findAllDebtors() {
        Query q = em.createQuery("select d from DebtorOne d");
        return (List<Debtor>) q.getResultList();
    }

then it works, but it is specific to DebtorOne schema...

One solution I see is to define a named query on the DebtorOne and DebtorTwo classes and call this named query from my DAO. In others words:

@Entity
@Table(name = "T_DEBTOR_ONE")
@NamedNativeQueries( { @NamedNativeQuery(name = "findAllDebtors", query = "select d from DebtorOne d") })
public class DebtorOne implements Debtor {

and in the DAO:

@SuppressWarnings("unchecked")
public List<Debtor> findAllDebtors() {
    Query q = em.createNamedQuery("findAllDebtors");
    return (List<Debtor>) q.getResultList();
}

I didn't try it yet, but I think it will work...

EDIT I just tried, this will work... except that the NamedQuery must be named differently for DebtorOne and DebtorTwo...

However, I am wondering if there is a way to solve my problem without using the latter solution?

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Edit regarding the first answer, which suggests to use the @MappedSuperclass. This annotation seems to be the perfect solution for me, but I think I forgot something, as I still get the same error.

The main Debtor:

@MappedSuperclass
public class Debtor {

    @Id
    @Column(name = "IDDEBTOR")
    protected String idDebtor; // With getter and setter

}

one of the extended Debtor class:

@Entity
@Table(name = "DEBTOR_ONE")
public class DebtorOne extends Debtor {

...

and in my DAO:

public List<Debtor> findAllDebtors() {
    return (List<Debtor>) em.createQuery("select d from Debtor d").getResultList();
}

is still returning me the error Caused by: org.hibernate.hql.ast.QuerySyntaxException: Debtor is not mapped [select d from Debtor d]

What did I miss this time?

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I think this is not possible with an interface, but only with a common abstract base class, that will be annotated with @MappedSuperclass (see Hibernate documentation for further details)

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I think for this to work you'd either have to actually map the Debtor table to a table or use the table per class strategy (union-subclass). @MappedSuperClass only seems to implement a very rudimentary mechanism to copy properties and wouldn't work because you can't query for instances of the superclass.

I take it from the link that you already have something in place to avoid mapping DebtorTwo in the hibernate session for the schema of DebtorOne (otherwise querying Debtor would pull in all records, including ones from the DebtorTwo table that wouldn't exist). In that case, follow the example from the documentation to map the subclass for either case.