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
I don’t believe you’ll be able to do that with the method name approach to query definition. From the documentation (reference): Although getting a query derived from the method name is quite convenient, one might face the situation in which either the method name parser does not support the keyword one wants to use or … Read more
Are you using Hibernate’s Query object, or JPA? For JPA, it should work fine: String jpql = “from A where name in (:names)”; Query q = em.createQuery(jpql); q.setParameter(“names”, l); For Hibernate’s, you’ll need to use the setParameterList: String hql = “from A where name in (:names)”; Query q = s.createQuery(hql); q.setParameterList(“names”, l);
Spring Data JPA 1.11 now supports the exists projection in repository query derivation. See documentation here. In your case the following will work: public interface MyEntityRepository extends CrudRepository<MyEntity, String> { boolean existsByFoo(String foo); }
If you do LIKE :code and then do namedQuery.setParameter(“code”, “%” + this.value + “%”); Then value remains free from the ‘%’ sign. If you need to use it somewhere else in the same query simply use another parameter name other than ‘code’ .
You are using JPQL which doesn’t support limiting results like this. When using native JPQL you should use setMaxResults to limit the results. However you are using Spring Data JPA which basically makes it pretty easy to do. See here in the reference guide on how to limit results based on a query. In your … Read more
I think you should use your (fully qualified) Status enum instead of literal value, so something like this: (assuming your Status enum is in com.myexample package) @NamedQuery(name = “Question.allApproved”, query = “SELECT q FROM Question q WHERE q.status = com.myexample.Status.APPROVED”).
I’m not sure for JPA 1.0 but you can pass a Collection in JPA 2.0: String qlString = “select item from Item item where item.name IN :names”; Query q = em.createQuery(qlString, Item.class); List<String> names = Arrays.asList(“foo”, “bar”); q.setParameter(“names”, names); List<Item> actual = q.getResultList(); assertNotNull(actual); assertEquals(2, actual.size()); Tested with EclipseLInk. With Hibernate 3.5.1, you’ll need to … Read more
When using IN with a collection-valued parameter you don’t need (…): @NamedQuery(name = “EventLog.viewDatesInclude”, query = “SELECT el FROM EventLog el WHERE el.timeMark >= :dateFrom AND ” + “el.timeMark <= :dateTo AND ” + “el.name IN :inclList”)