如何創建標準API查詢

Question

我有一個類：

@Entity 
public class Resume { 
    private Long id; 
    @Embedded 
    private DesiredPositionAndSalary desiredPositionAndSalary; 
} 

和類：

@Embeddable 
public class DesiredPositionAndSalary { 
    @ManyToMany 
    private Set<Specialization> specializations; 
} 

和類;）

@Entity 
public class Specialization { 
    private Long id; 
} 

現在我有一些專業化，我需要過濾。 例如，我需要選擇所有的專業化程序員或經理之一的簡歷。類似於

select * from resume r inner join resume_to_specialization rts on r.id = rts.id inner join specialization s on rts.spec_id in(1,2) 

我該如何在Criteria API中編寫此查詢？如果我想念一些重要細節，我可以給予更多。

Answer 1

好吧，我這個處理：

CriteriaBuilder cb = entityManager.getCriteriaBuilder(); 
CriteriaQuery<Resume> cq =cb.createQuery(Resume.class); 
Root<Resume> root = cq.from(Resume.class); 
cq.select(root); 

Set<Specialization> filter = getFilter(); 

SetJoin<DesiredPositionAndSalary, Specialization> join = root.join(Resume_.desiredPositionAndSalary, JoinType.INNER).join(
      DesiredPositionAndSalary_.specializations, JoinType.INNER); 

cq.where(cb.and(cq.getRestriction(), join.in(filter))); 
cq.distinct(true);/*it is major, or we get duplicate of resume for every 
        specialization overlap with filter*/ 

List<Resume> result = em.createQuery(cq).getResultList(); 

。