如何刪除刪除連接表實體JPA 2.1 CriteriaDelete

Question

我想刪除（JPA 2.1）所有的「患者」，從一個「醫院」，但碰到一個問題： 「UPDATE/DELETE標準查詢不能定義連接」

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(); 

錯誤：

UPDATE/DELETE criteria queries cannot define joins 

我應該如何刪除所有的患者，而加入不能執行？

Answer 1

您可以使用選擇「在」條款適當的實體和一個子查詢。

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();