HQL查詢幫助

Question

我有問題編寫HQL以顯示不同的applicationId與最新的應用程序（newest createDate）爲表中的以下數據。

+---------------+-------------+----------+---------------------+ 
| applicationId | firstName | lastName | createDate   | 
+---------------+-------------+----------+---------------------+ 
|    1 | Mark  | Johnson | 2010-05-03 00:00:00 | 
|    3 | Henry  | Jordan | 2010-05-03 00:00:00 | 
|    5 | Cindy Spahn | Wilson | 2010-05-03 00:00:00 | 
|    5 | Cindy Spahn | Wilson | 2010-05-04 00:00:00 | 
|    5 | Cindy Spahn | Wilson | 2010-05-05 00:00:00 | 
+---------------+-------------+----------+---------------------+ 
5 rows in set (0.00 sec) 

下面是我在尋找的結果：

+---------------+-------------+----------+---------------------+ 
| applicationId | firstName | lastName | createDate   | 
+---------------+-------------+----------+---------------------+ 
|    1 | Mark  | Johnson | 2010-05-03 00:00:00 | 
|    3 | Henry  | Jordan | 2010-05-03 00:00:00 | 
|    5 | Cindy Spahn | Wilson | 2010-05-05 00:00:00 | 
+---------------+-------------+----------+---------------------+ 
3 rows in set (0.00 sec) 

實體如下：

@Entity 
@Table(name = "application") 
public class Application { 
    private long applicationId; 
    private String firstName; 
    private String lastName; 
    private List<ApplicationHistory> applicationHistoryList; 

    @Id 
    @GeneratedValue(strategy = GenerationType.AUTO) 
    public long getApplicationId() { 
     return applicationId; 
    } 

    @OneToMany(mappedBy = "application", fetch = FetchType.EAGER, cascade = CascadeType.ALL) 
    public List<ApplicationHistory> getApplicationHistoryList() { 
     return applicationHistoryList; 
    } 
    // getter() and setter() 
} 

和：

@Entity 
@Table(name = "applicationHistory") 
public class ApplicationHistory { 
    private Application application; 
    private final Timestamp createDate = new Timestamp(System.currentTimeMillis()); 

    @ManyToOne 
    @JoinColumn(name = "applicationId", insertable = false, updatable = false) 
    public Application getApplication() { 
     return application; 
    } 

    @Id 
    @Column(columnDefinition = "timestamp default current_timestamp") 
    public Timestamp getCreateDate() { 
     return createDate; 
    } 
} 

Answer 1

何樂而不爲呢與分組條款：

select ah from ApplicationHistory ah group by ah.applicationId order by ah.createDate desc 

Answer 2

您可以使用下面的查詢：

select a, h.createDate from Application as a join a.applicationHistoryList as h where (a.applicationId, h.createDate) in(SELECT application.applicationId, max(createDate) FROM ApplicationHistory group by application.applicationId) 

如：

Query q = em.createQuery("select a, h.createDate from Application as a join a.applicationHistoryList as h where (a.applicationId, h.createDate) in(SELECT application.applicationId, max(createDate) FROM ApplicationHistory group by application.applicationId) "); 

     List list = q.getResultList(); 

     for (Iterator iterator = list.iterator(); iterator.hasNext();) { 
      Object obj[] = (Object[])iterator.next(); 
      Application a = (Application) obj[0]; 

      System.out.println("ApplicationId="+a.getApplicationId()); 
      System.out.println("CreateDate="+obj[1]); 

     }