使用条件生成器运行 IN 查询子句

Question

我需要一些帮助来运行下面的查询，使用 CriteriaBuilder aproach：

select count(*) from TABLE_VIEW_SEARCH where
        person_ID in (select person_ID from TABLE_PERSON
        where region_ID = 1001) and postal_cd ='AL';

我试过这个，但它失败了：

    CriteriaBuilder cb = ..;
    CriteriaQuery<Long> criteriaQuery= cb.createQuery(Long.class);
    Root<ViewSearchEntity> from= criteriaQuery.from(ViewSearchEntity.class);
    Subquery<Long> subQuery = criteriaQuery.subquery(Long.class);
    Root<PersonEntity> subFrom = subQuery .from(PersonEntity.class);
    subQuery .where(cb.equal(subFrom.get(PersonEntity_.region_ID ), region_ID ));
    criteriaQuery.select(cb.count(from));
    criteriaQuery.where(meFrom.in(subQuery );
    TypedQuery<Long> query = getEntityManager().createQuery(criteriaQuery);
    Long result = query.getSingleResult();

我

达成了一个解决方案，基本上我设法做到了连接条件：

    CriteriaBuilder cb = ..;
    CriteriaQuery<Long> criteriaQuery= cb.createQuery(Long.class);
    Root<ViewSearchEntity> from= criteriaQuery.from(ViewSearchEntity.class);
    Predicate predicate1 = cb.equal(from.get(ViewSearchEntity.postal_cd),"AL");
    Subquery<Long> subQuery = criteriaQuery.subquery(Long.class);
    Root<PersonEntity> subRoot = subQuery.from(PersonEntity.class);
    Join<PersonEntity, RegionEntity_> region= subRoot.join(RegionEntity_.region);
    subQuery.select(region.get(RegionEntity_.regionId)).where(
            cb.equal(subRoot.get(PersonEntity_.region), regionId));
    Predicate predicateInClause = root.get(ViewSearchEntity_.postal_cd).in(subQuery);
    return getEntityManager().createQuery(criteriaQuery.select(cb.count(root)).where(predicate1,predicateInClause)).getEntityManager().createQuery(criteriaQuery.select(cb.count(root)).where(predicate1,predicateInClause))

Answer 1

像这样的东西。由于缺少实体名称，我无法给您查询。但是，我建议阅读此标准 API 子选择

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery cq = cb.createQuery(TABLE_VIEW_SEARCH.class);
Root root = cq.from(TABLE_VIEW_SEARCH.class);
Subquery sub = cq.subquery(TABLE_PERSON.class);
Root subRoot = sub.from(TABLE_PERSON.class);
SetJoin<TABLE_PERSON, TABLE_VIEW_SEARCH> subViewSearch = 
     subRoot.join(TABLE_PERSON.tableViewSearch);
sub.select(cb.equal(subRoot.get(TABLE_PERSON.region_ID), 1001));
cq.where(cb.equal(root.get("postal_cd"), "AL"));
TypedQuery query = em.createQuery(cq);
List result = query.getResultList();

声明元模型生成依赖项后，可以使用此方法：

<dependency>
    <groupId>org.hibernate</groupId>
    <artifactId>hibernate-jpamodelgen</artifactId>
</dependency>

添加此依赖项后，使用 maven 重新生成项目，然后可以执行以下操作：

EntityName_.attribute