如何在JPA critera API中进行不同计数?

问题描述 投票:11回答:4

我想这样做,但要使用标准API:

select count(distinct e) from Event e, IN(e.userAccessPermissions) p where p.principal = :principal and p.permission in (:permissions)

有什么想法吗?

java hibernate jpa criteria-api
4个回答
24
投票

您可以在CriteriaBuilder上使用countDistinct

criteriaQuery.select(criteriaBuilder.countDistinct(entityRoot))

1
投票

喜欢吗?

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"));

0
投票

在查询上使用c.distinct(true)。

请参见http://relation.to/Bloggers/ATypesafeCriteriaQueryAPIForJPA以获取更多示例。


0
投票
    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();


    }

使用Spring Data Jpa,我们可以使用此方法:

     /*
     * (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()));
    }
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