querydsl从元组获取列

Question

I've created query that returns building id and apartments count in that building. How to get buildings with max count from this tuple? Other than iterate through entire list.

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);

Here are building and apartment entities:

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;
}

Answer 1

You could order the query by apartment.count() in descending order and pick the first result. Something like:

query2.select(apartment.count(), apartment.building.id)
      .from(apartment)
      .where(apartment.status.lower().eq("free"))
      .groupBy(apartment.building)
      .orderBy(apartment.count().desc())
      .limit(1L);