【发布时间】:2017-09-12 08:31:18
【问题描述】:
我想使用 JPA 2.0 CriteriaBuilder 创建以下 SQL 查询:
SELECT * FROM PROFIL WHERE PROFILID IN (SELECT PROFILID FROM ROLEPROFIL WHERE ROLEID = roleId)
这是我的课程:
简介:
public class Profil extends AbstractDomain<Long> {
@Id
@Column
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "idgen_x")
@UiInfo(name = "Identifiant")
private Long profilId;
@Column(nullable = false)
@UiInfo(name = "Libellé")
private String lib;
}
角色:
public class Role extends AbstractDomain<Long> {
@Id
@Column
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "idgen_x")
@UiInfo(name = "Identifiant")
private Long roleId;
}
角色简介:
public class RoleProfil extends AbstractDomain<Long> {
@Id
@Column
@UiInfo(name = "Identifiant")
private Long roleProfilId;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(nullable = false)
@UiInfo(name = "Profil")
private Profil profilId;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(nullable = false)
@UiInfo(name = "Rôle")
private Role roleId;
}
我想做的是创建一个函数,该函数将使用 JPA 通过 Role ID 获取 Profils。
这是我开始的,但我是 JPA 的新手,我不知道该怎么做:
public List<Profil> findProfilsByRoleId(Long roleId) {
final CriteriaBuilder builder = getCriteriaBuilder();
final CriteriaQuery<Profil> criteriaQuery = builder.createQuery(Profil.class);
final Root<Profil> from = criteriaQuery.from(Profil.class);
//TODO
TypedQuery<Profil> query = getEntityManager().createQuery(criteriaQuery);
return query.getResultList();
}
我该怎么做?
【问题讨论】: