Is there a way to populate non-column member using a query in Spring JPARepository?

Question

I have the following structure:

@MappedSuperclass
public abstract class Action {
...
}

And than few entities extending Action, for example:

@Entity(name="chemistry")
public class Chemistry extends Action implements Serializable {
...
}

In order to execute the persistance I used JPARepository:

@NoRepositoryBean
public interface ActionRepository<T extends Action> extends JpaRepository<T, String> {
...
}

And an interface for the children:

@Component 
@Service
@Repository
interface ChemistryRepository extends ActionRepository<Chemistry> {
}

The persistance mechanism works fine for all the entities inheriting from Action. Now I have a problem. I have another table in the DB, called 'action_params', including additional key-value parameters which related to a specific action (ie chemistry). I have the appropriate entity:

@Entity(name="action_params")
public class ActionParams implements Serializable {
...
}

In the DB I cannot use foreign key since the key is not related to a specific table but any table inherits from Action. this is ok, I would like to use a query in order to get for each Action the records from the table action_params, based on its id.

And I would like to add a List to Action, mapping those records as a list. I tried using hibernate but any query returns empty (The following code was added into Action class):

@OneToMany
@JoinFormula(value="select p from action_params p where p.action_id = id")
@Transient
protected List<ActionParams>  actionParamsList;

This is not working. I also tried to autowire the ActionParamsRepository into Action, but of course there is no way doing it and I could not think of any other way.

Your advice will be much appreciated

Answer 1

What strong reason you have to use a JPA relation?

If there is no such reason, I'd suggest you to implement a new finder method in the ActionParams repository. Something like this:

@Query("select p from action_params p where p.action_id = :actionId")
List<ActionParams> findAllByActionId(@Param("actionId") Long actionId )

I suppose the type is Long. Replace it if you have other type.

Answer 2

You can use @Formula , so let me explain, when you put a query on the formula, in Hibernate is added a subquery.

@Formula("(SELECT p FROM action_params p WHERE p.action_id = id)")
private List<ActionParams>  actionParamsList;