如何在已加入的实体上调用订单?我试图通过以下方式实现以下目标:
从p.id = t.person_id上的人员p内联接电话t中选择*通过s.name DESC在s.id = t.sim_id上加入sim s
@Entity
public class Person implements Serializable{
@Id
private Long id;
@OneToMany(orphanRemoval = true, mappedBy = "person", fetch = FetchType.LAZY, cascade = CascadeType.PERSIST)
private List<Telephone> telephonesNumber;
@Entity
public class Telephone implements Serializable {
@Id
private String number;
@Id
@ManyToOne()
@JoinColumn(name = "person_id")
private Person person;
@Id
@ManyToOne(cascade = {})
@JoinColumn(name = "sim_id")
private Sim sim;
@Entity
public class Sim implements Serializable {
@Id
private Long id;
@Column(unique = true)
private String name;
我使用规范界面,在这个例子中,排序是在字段person.id上并且它可以正常工作
public class PersonSpecification implements Specification<Person> {
@Override
public Predicate toPredicate(Root<Person> root, CriteriaQuery<?> query, CriteriaBuilder builder) {
List<Predicate> predicates = new ArrayList<>();
// there is many different conditions for example
// if(someCondition!=null) {
// predicates.add(builder.like(root.get("someProperty"), someValue));
// }
query.groupBy(root.get("id"));
//there I want to order by Sim.name i dont know how
query.orderBy(builder.asc(root.get("phone")));//this works
return builder.and((predicates.toArray(new Predicate[predicates.size()])));
}
我想通过Sim.name订购,但我不知道如何。
答案 0 :(得分:5)
在JPA规范中,您可以使用:
query.orderBy(builder.asc(root.join("telephonesNumber").get("sim").get("name")));
按SIM名称排序。
如果您使用JPA查询:
@Query("select s from Person p
join p.telephonesNumber t
join t.sim s order
by t.sim.id desc")
它会产生这个:
select * from person p
inner join telephone t on p.id=t.person_id
inner join sim s on t.sim_id=s.id
order by t.sim_id desc
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