[英]querydsl get column from tuple
我创建了返回建筑物ID和该建筑物中公寓数量的查询。 如何从该元组中获得具有最大数量的建筑物? 除了遍历整个列表。
JPAQuery<Tuple> query2 = new JPAQuery(entityManager);
QBuildingEntity building = QBuildingEntity.buildingEntity;
QApartmentEntity apartment = QApartmentEntity.apartmentEntity;
query2 = query2.select(apartment.count(), apartment.building.id).from(apartment).where(apartment.status.lower().eq("free"))
.groupBy(apartment.building);
以下是建筑物和公寓实体:
public class BuildingEntity extends AbstractEntity implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
Long id;
@OneToMany(fetch = FetchType.LAZY, mappedBy = "building", cascade = CascadeType.ALL, orphanRemoval = true)
Set<ApartmentEntity> apartments = new HashSet<>();
}
public class ApartmentEntity extends AbstractEntity implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
Long id;
@Column(name = "STATUS", nullable = false, length = 50)
String status;
@ManyToOne(fetch = FetchType.LAZY, cascade = {CascadeType.PERSIST, CascadeType.MERGE, CascadeType.REFRESH})
@JoinColumn(name = "BUILDING_ID")
BuildingEntity building;
}
您可以通过apartment.count()
降序对查询进行排序,然后选择第一个结果。 就像是:
query2.select(apartment.count(), apartment.building.id)
.from(apartment)
.where(apartment.status.lower().eq("free"))
.groupBy(apartment.building)
.orderBy(apartment.count().desc())
.limit(1L);
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