[英]How to do a distinct count in JPA critera API?
I would like to do this but with the criteria API instead:我想这样做,但使用标准 API 代替:
select count(distinct e) from Event e, IN(e.userAccessPermissions) p where p.principal = :principal and p.permission in (:permissions)
Any ideas?有任何想法吗?
You can use countDistinct on CriteriaBuilder您可以在 CriteriaBuilder 上使用 countDistinct
criteriaQuery.select(criteriaBuilder.countDistinct(entityRoot))
public long getCount(String xValue){
EntityManager entityManager = this.getEntityManager();
CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Long> criteriaQuery = cb.createQuery(Long.class);
Root<MyEntity> root = criteriaQuery.from(MyEntity.class);
criteriaQuery.select(cb.count(criteriaQuery.from(MyEntity.class)));
List<Predicate> predicates = new ArrayList<>();
Predicate xEquals = cb.equal(root.get("x"), xValue);
predicates.add(xEquals);
criteriaQuery.select(cb.countDistinct(root));
criteriaQuery.where(predicates.toArray(new Predicate[0]));
return entityManager.createQuery(criteriaQuery).getSingleResult();
}
With Spring Data Jpa, we can use this method:使用 Spring 数据 Jpa,我们可以使用这种方法:
/*
* (non-Javadoc)
* @see org.springframework.data.jpa.repository.JpaSpecificationExecutor#count(org.springframework.data.jpa.domain.Specification)
*/
@Override
public long count(@Nullable Specification<T> spec) {
return executeCountQuery(getCountQuery(spec, getDomainClass()));
}
Use c.distinct(true) on your Query.在查询中使用 c.distinct(true)。
See http://relation.to/Bloggers/ATypesafeCriteriaQueryAPIForJPA for more samples.有关更多示例,请参阅http://relation.to/Bloggers/ATypesafeCriteriaQueryAPIForJPA 。
Like this?像这样?
Criteria crit = session.createCriteria(Event.class):
crit.createAlias("userAccessPermissions", "p");
crit.add(Restrictions.eq("p.principal", principal);
crit.add(Restrictions.in("p.permission", permissions);
crit.setProjection(Projections.countDistinct("id"));
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