Question

这里我使用NativeQuery来执行使用子查询选择lookuptype这是正常的，但我想使用Criteria Builder我可以如何使用

Query query = em.createNativeQuery("SELECT * FROM LOOKUPMASTER WHERE PARENTLOOKUPTYPEID = (SELECT LOOKUPID FROM LOOKUPMASTER WHERE LOOKUPTYPE =? ) ", Lookupmaster.class); query.setParameter(1, lookUpType);

我尝试使用条件构建器编写上述查询，但我得到的结果不同，这里是我的条件查询

`CriteriaBuilder cb =em.getCriteriaBuilder();
        CriteriaQuery cq=cb.createQuery(Lookupmaster.class);
        Root<Lookupmaster> rt=cq.from(Lookupmaster.class);
        Path<Object> path = rt.get("parentlookuptypeid");
        cq.select(rt);
        Subquery<Lookupmaster> subquery = cq.subquery(Lookupmaster.class);
        Root rt1 = subquery.from(Lookupmaster.class);
        subquery.select(rt1.get("lookupid"));
        subquery.where(cb.equal(rt.get("lookuptype"),lookUpType));

        cq.where(cb.in(path).value(subquery));
        Query qry =em.createQuery(cq)`

Answer 1

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.equal("name",name_value));
subquery.where(criteriaBuilder.equal("id",id_value));

select.where(criteriaBuilder.in(path).value(subquery));

TypedQuery<Object> typedQuery = entityManager.createQuery(select);
List<Object> resultList = typedQuery.getResultList();

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如何使用条件构建器编写子查询

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