I've got a JPA @MappedSuperClass
and an @Entity
extending it:
@MappedSuperclass
public class BaseClass {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
@Column
private Boolean active;
//getters & setters
}
@Entity
public class Worker extends BaseClass{
@Column
private String name;
//getters & setters
}
The active
field of the base class is a flag for the children entities. Only the active ones should be loaded in the application. Then I've written a generic Spring Data Proxy interface:
public interface Dao<T extends BaseClass, E extends Serializable> extends
CrudRepository<T, E> {
Iterable<T> findByActive(Boolean active);
}
And this one is the interface that should be for Worker
data access, properly extending the previous one:
@Transactional
public interface WorkerDao extends Dao<Worker, Long>{}
Well, now in my logic layer I've implemented an abstract class which will wrap the common code for CRUD operations over my entities. I'll have a service for each of them, but I want just to inherit from the abstract
one. I want to wire the specific repository for each of the services and provide it to the superclass using an abstract
method. That's how my superclass is implemented:
public abstract class GenericService<E extends BaseClass>{
public abstract Dao<E, Long> getDao();
//Here I've got some common operations for managing
//all my application classes, including Worker
}
The problem is that the getDao()
method uses the E
class parameter, which is guaranteed only to be a child of BaseClass
and not a javax.persistence.Entity
. When I try to access the DAO from my custom service implementation I get this error:
Caused by: java.lang.IllegalArgumentException: Could not create query metamodel for method public abstract java.lang.Iterable com.mycompany.model.daos.interfaces.Dao.findByActive(java.lang.Boolean)!
at org.springframework.data.jpa.repository.query.JpaQueryLookupStrategy$CreateQueryLookupStrategy.resolveQuery(JpaQueryLookupStrategy.java:93)
Caused by: java.lang.IllegalArgumentException: Not an entity: class com.mycompany.model.BaseClass
at org.hibernate.jpa.internal.metamodel.MetamodelImpl.entity(MetamodelImpl.java:203)
Which makes sense, because E
is defined as a child of BaseClass
. The compiler allows me to write this too:
public abstract class GenericService<E extends BaseClass && Entity>
However I get an error in the child Service that says Worker
class is not compatible with the signature for E
. Does anybody know how to solve this?
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