帶有聯結表的JPA CriteriaBuilder

Question

下面的sql表達式如何使用CriteriaBuilder表示？

select * from Ref where prac_id = (select prac_id from loc l join staff_loc sl where sl.loc = l.id and sl.pracstaff_id = 123)

模型類

@Entity
public class Ref {
    private Long id;
    private Prac prac;
}

@Entity
public class Loc {
    Long id;
    @ManyToOne
    Prac prac;
    @ManyToMany
    Set<PracStaff> pracStaff;
}

@Entity
public class Prac {
    Long id;
    @OneToMany
    Set<Loc> locs;
}

@Entity
public class PracStaff {
    Long id;
    @ManyToMany
    Set<Loc> locs;
}

• 有一個聯接表，將Loc映射到PracStaff。 它有兩列：pracstaff_id和loc_id
• Loc只能屬於一個Prac。

我想要獲取的是使用CriteriaBuilder具有ID為123的PracStaff的所有Ref對象。

Answer 1

---

這是我必須進行全面測試的解決方案。 運用

Expression<Collection<PracStaff>>

返回收藏是我所缺少的

CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<Ref> criteriaQuery = criteriaBuilder.createQuery(Ref.class);
Root<Ref> from = criteriaQuery.from(Ref.class);
criteriaQuery.select(from);

Subquery<Prac> subquery = criteriaQuery.subquery(Prac.class);
Root<Loc> fromLoc = subquery.from(Loc.class);
Expression<Collection<PracStaff>> pracStaffInLoc = fromLoc.get("pracStaff");
subquery.where(criteriaBuilder.isMember({pracStaffObj}, pracStaffInLoc));
subquery.select(fromLoc.<Prac>get("prac"));     
Path<Prac> specialist = from.get("{field in Ref class}");
Predicate p = criteriaBuilder.equal(specialist, subquery);