Maybe the following extract from the Chapter 23 – Using the Criteria API to Create Queries of the Java EE 6 tutorial will throw some light (actually, I suggest reading the whole Chapter 23): Querying Relationships Using Joins For queries that navigate to related entity classes, the query must define a join to the related … Read more
If I understand well, you want to Join ScheduleRequest with User and apply the in clause to the userName property of the entity User. I’d need to work a bit on this schema. But you can try with this trick, that is much more readable than the code you posted, and avoids the Join part … Read more
One of the JPA ways for getting only particular columns is to ask for a Tuple object. In your case you would need to write something like this: CriteriaQuery<Tuple> cq = builder.createTupleQuery(); // write the Root, Path elements as usual Root<EntityClazz> root = cq.from(EntityClazz.class); cq.multiselect(root.get(EntityClazz_.ID), root.get(EntityClazz_.VERSION)); //using metamodel List<Tuple> tupleResult = em.createQuery(cq).getResultList(); for (Tuple t … Read more
Below is the pseudo-code for using sub-query using Criteria API. CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder(); CriteriaQuery<Object> criteriaQuery = criteriaBuilder.createQuery(); Root<EMPLOYEE> from = criteriaQuery.from(EMPLOYEE.class); Path<Object> path = from.get(“compare_field”); // field to map with sub-query from.fetch(“name”); from.fetch(“id”); CriteriaQuery<Object> select = criteriaQuery.select(from); Subquery<PROJECT> subquery = criteriaQuery.subquery(PROJECT.class); Root fromProject = subquery.from(PROJECT.class); subquery.select(fromProject.get(“requiredColumnName”)); // field to map with main-query subquery.where(criteriaBuilder.and(criteriaBuilder.equal(“name”,name_value),criteriaBuilder.equal(“id”,id_value))); select.where(criteriaBuilder.in(path).value(subquery)); … Read more