Why not using the @Query annotation? @Query(“select p.id from #{#entityName} p”) List<Long> getAllIds(); The only disadvantage I see is when the attribute id changes, but since this is a very common name and unlikely to change (id = primary key), this should be ok.
Perhaps you need criteria.add(cb.like(emp.<String>get(“name”), p)); because first argument of like() is Expression<String>, not Expression<?> as in equal(). Another approach is to enable generation of the static metamodel (see docs of your JPA implementation) and use typesafe Criteria API: criteria.add(cb.like(emp.get(Employee_.name), p)); (Note that you can’t get static metamodel from em.getMetamodel(), you need to generate it by … Read more
The problem is that with the string-based API it cannot infer the type for the result value of the get-Operation. This is explained for example in Javadoc for Path. If you use predicates.add(builder.lessThanOrEqualTo(root.<Date>get(“dateCreated”), param)); instead, it will work fine, because it can figure out the return type from the type argument and will find out … 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
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
If you need to add couple of orders you can make something like (but for your query and different root objects) CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder(); CriteriaQuery<Route> query = criteriaBuilder.createQuery(Route.class); Root<Route> routeRoot = query.from(Route.class); query.select(routeRoot); List<Order> orderList = new ArrayList(); query.where(routeRoot.get(“owner”).in(user)); orderList.add(criteriaBuilder.desc(routeRoot.get(“date”))); orderList.add(criteriaBuilder.desc(routeRoot.get(“rating”))); query.orderBy(orderList);
I’m pretty sure this has already been covered here on SO but I couldn’t find the existing question. So, here is my point of view on the question: I find JPQL queries easier to write/read. I find the Criteria API nice for building dynamic queries. Which is basically what you’ll find in Hibernate: Criteria vs. … 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