Question

我是jpql的新手，我有以下情况。

我有两个实体地点和地址。

@Entity
public class Place{


   @OneToMany
   private List<Address> addresses;

   ....
}

 @Entity
 public class Address{

   String description; 

   Date dataFrom;

   Date dataTo;

   @ManyToOne
   private Place place;


   ....
}

我想获得最后一个地址的描述。我正在尝试这样做：

select a.description from place p join p.addresses a.....

现在我应该按时间顺序得到最后一个地址。我该怎么办？

Answer 1

SELECT addresses.description 
FROM place p JOIN p.addresses addresses 
ORDER BY addresses.dateFrom

然后将此作为resultList返回并获取列表中的第一项，我会说你可以在T-SQL SELECT TOP 1中做到，但是，我不相信JPQL支持这个。 / p>

Answer 2

尝试使用子查询，例如

select a.description from place p join p.addresses a where a.dataFrom = (select max(address.dataFrom) from Address address where address.place = p)

Answer 3

  Criteria criteria = session.createCriteria(Place.class, "place"); 
  criteria.createAlias("place.addresses", "addresses")
  criteria.add(Order.desc("addresses.dataFrom"));
  criteria.setProjection(Projections.property("addresses"));

  List<Address> addresses = criteria.list();

  for (Address address : addresses) {
       System.out.println(address.description);
  }

如果您想根据地点ID使用以下地址找到地址。

  Criteria criteria = session.createCriteria(Place.class, "place"); 
  criteria.add(Restrictions.eq("place.id", put the place id here);
  criteria.createAlias("place.addresses", "addresses")
  criteria.add(Order.desc("addresses.dataFrom"));
  criteria.setProjection(Projections.property("addresses"));

  List<Address> addresses = criteria.list();

  for (Address address : addresses) {
       System.out.println(address.description);
  }

一对多查询jpql

3 个答案: