Polymorphic JPA query with CriteriaQuery API

Question

I have the following entity structure:

   +-----------+                +-------------+        
   |  User     | -------------> |    Role     |
   +-------         


        

Answer 1

Solved it using a subquery

I created a subquery based on the Role2 type, joined it with the "SomeEntity" entity and applied the predicates. Then I connected the subquery to the "main" query using the Role2.ids that match the subquery predicates.

CriteriaBuilder cb = ...
CriteriaQuery<String> query = cb.createQuery(String.class);
Root<User> user = query.from(User.class);
SetJoin<User, Role> userRolesJoin = user.join(User_.roles);
Path<String> roleIdPath = userRolesJoin.get(User_.id);

Subquery<String> subquery = query.subquery(String.class);
Root<Role2> role2Root = subquery.from(Role2.class);
Join<Role2, SomeEntity> someEntityJoin = role2Root.join(Role2_.someEntity);
Path<String> someEntityPropertyPath = someEntityJoin.get(SomeEntity_.aProperty);
Predicate someEntityPropertyPredicate = cb.equal(someEntityPropertyPath,
            "a property value");
subquery.select(role2Root.get(Role2_.id));
subquery.where(someEntityPropertyPredicate);

In<String> idInSubqueryIds = cb.in(roleIdPath).value(subquery);

query.select(user.get(User_.username));
query.where(idInSubqueryIds);

EntityManager entityManager = ...;
TypedQuery<String> query = entityManager.createQuery(query);
List<String> usernames = query.getResultList();

Answer 2

You can use as() in Criteria to cast.

((Path)user.join(User_.roles).as(Role2.class)).join(Role2_.someEntity)

This may be provider specific previous to JPA 2.1, but I think it is standard in JPA 2.1.

In JPQL you can use the TREAT operation.

http://wiki.eclipse.org/EclipseLink/UserGuide/JPA/Basic_JPA_Development/Querying/JPQL#Special_Operators

You can also use a second from(),

Root<Role2> role2 = query.from(Role2.class);
... cb.equal(user.join(User_.roles), role2)