如何使用JPA 2.1 CriteriaDelete从联接表中删除删除实体

Question

我想从一个“医院”中删除（JPA 2.1）所有“患者”，但是遇到一个问题： “ UPDATE / DELETE条件查询无法定义联接”

CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaDelete<PatientEntity> delete = cb.createCriteriaDelete(PatientEntity.class);
Root<PatientEntity> root = delete.from(PatientEntity.class);
Join<PatientEntity, HospitalEntity> join = root.join(PatientEntity_.Hospital);
delete.where(cb.equal(join.get(HospitalEntity_.id), id));
Query query = entityManager.createQuery(delete);
query.executeUpdate();

错误：

UPDATE/DELETE criteria queries cannot define joins

无法执行加入时，我应该如何删除所有患者？

Answer 1

您可以使用子查询来选择适当的实体并为此选择“ in”子句。

CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaDelete<PatientEntity> delete = cb.createCriteriaDelete(PatientEntity.class);
Root<PatientEntity> root = delete.from(PatientEntity.class);


                Subquery<PatientEntity> subquery = delete.subquery(PatientEntity.class);
                Root<PatientEntity> root2 = subquery.from(PatientEntity.class);
                subquery.select(root2);
                /* below are narrowing criteria, based on root2*/   
                Join<PatientEntity, HospitalEntity> join = root2.join(PatientEntity_.Hospital);
                subquery.where(cb.equal(join.get(HospitalEntity_.id), id));


delete.where(root.in(subquery));
Query query = entityManager.createQuery(delete);
query.executeUpdate();