9

在 SpringBoot 微服务中，我试图为每个 mean_of_payment_id 选择一个演员的最新记录。为此，使用 mean_of_payment_id 上的 group by 子句为 actor_id 选择演员内容，其中 created_date 等于 max(created_date) 的嵌套查询的子集。我正在使用 JPQL。下面是表结构和查询。

在此处输入图像描述

    @Query("select ac from ActorContent ac "
        + "where (ac.actor.uuid=:actorUuid ) and "
        + "ac.createdDate IN ( SELECT MAX(aci.createdDate) "
            + "FROM ActorContent aci WHERE ac.actor.uuid=aci.actor.uuid "
            + "and aci.uuid = ac.uuid group by ac.meanOfPayment.id)"
        )

在此处输入图像描述

不幸的是，执行查询后，我得到了所有记录，但我期望的是前三行。MeanOfPayment 和 Actor 是 ActorContent 的参考表。

4

1 回答 1

4

我认为就关系代数而言，您要求的集合ActorContent 减去ActorContent受actor = actor 和 meanOfPayment = meanOfPayment 和 createDate < createDate 约束的集合。ActorContent因此，考虑它的方法是从with的叉积中获得第二组ac1.meanOfPayment = ac2.meanOfPayment and ac1.actor = ac2.actor and ac1.createDate < ac2.createDate。然后从集合中减去这个集合ActorContent。我还没有查看它是否比使用MAX和Group By通过示例更有效：

@Query("select ac from ActorContent ac where ac.id not in (select ac1.id from ActorContent ac1, ActorContent ac2 where ac1.meanOfPayment = ac2.meanOfPayment and ac1.actor = ac2.actor and ac1.createDate < ac2.createDate)")

这给了我 UPPER 表中的前四行，代表第一个演员和他唯一的 meanOfPayment，第二个演员和他最近对所有三个 meanOfPayment 的付款。

ActorContent [id=1, actor=Actor [id=1], meanOfPayment=MeanOfPayment [id=1], amount=10500.00, createDate=2018-10-09 00:00:00.887]
ActorContent [id=2, actor=Actor [id=2], meanOfPayment=MeanOfPayment [id=1], amount=-10400.00, createDate=2018-10-02 00:00:00.887]
ActorContent [id=3, actor=Actor [id=2], meanOfPayment=MeanOfPayment [id=3], amount=6000.00, createDate=2018-10-02 00:00:00.887]
ActorContent [id=4, actor=Actor [id=2], meanOfPayment=MeanOfPayment [id=2], amount=200.00, createDate=2018-09-30 00:00:00.887]

之后，您可能希望通过连接获取Actor和MeanOfPayment实例来优化查询。举例：

@Query("select ac from ActorContent ac left outer join fetch ac.actor left outer join fetch ac.meanOfPayment where ac.id not in (select ac1.id from ActorContent ac1, ActorContent ac2 where ac1.meanOfPayment = ac2.meanOfPayment and ac1.actor = ac2.actor and ac1.createDate < ac2.createDate)")

这将导致以下休眠生成的 SQL 查询：

select actorconte0_.id as id1_1_0_, actor1_.id as id1_0_1_, meanofpaym2_.id as id1_2_2_, actorconte0_.actor_id as actor_id4_1_0_, actorconte0_.amount as amount2_1_0_, actorconte0_.create_date as create_d3_1_0_, actorconte0_.mean_of_payment_id as mean_of_5_1_0_ from actor_content actorconte0_ left outer join actor actor1_ on actorconte0_.actor_id=actor1_.id left outer join mean_of_payment meanofpaym2_ on actorconte0_.mean_of_payment_id=meanofpaym2_.id where actorconte0_.id not in  (select actorconte3_.id from actor_content actorconte3_ cross join actor_content actorconte4_ where actorconte3_.mean_of_payment_id=actorconte4_.mean_of_payment_id and actorconte3_.actor_id=actorconte4_.actor_id and actorconte3_.create_date<actorconte4_.create_date)

当然，如果您想要一个特定Actor的，那么只需将其添加到 where 子句中。

@Query("select ac from ActorContent ac left outer join fetch ac.actor left outer join fetch ac.meanOfPayment where ac.actor.id = :actorId and ac.id not in (select ac1.id from ActorContent ac1, ActorContent ac2 where ac1.meanOfPayment = ac2.meanOfPayment and ac1.actor = ac2.actor and ac1.createDate < ac2.createDate)")
public List<ActorContent> findLatestForActor(@Param("actorId") Integer actorId);

这给了我“前三行”

ActorContent [id=2, actor=Actor [id=2], meanOfPayment=MeanOfPayment [id=1], amount=-10400.00, createDate=2018-10-02 00:00:00.066]
ActorContent [id=3, actor=Actor [id=2], meanOfPayment=MeanOfPayment [id=3], amount=6000.00, createDate=2018-10-02 00:00:00.066]
ActorContent [id=4, actor=Actor [id=2], meanOfPayment=MeanOfPayment [id=2], amount=200.00, createDate=2018-09-30 00:00:00.066]

如果您对 Actor 和 MeanOfPayment 组合具有相同的 createDate 有疑问，那么您可以通过几种不同的方式进行处理。首先，如果您有一个逻辑约束，以至于您不想处理这些重复项，那么您可能也应该有一个数据库约束，这样您就不会得到它们，并确保您不会首先创建它们。另一件事是您可以手动检查结果列表并将其删除。最后，您可以在查询中使用 distinct，但您必须省略ActorContentid 字段，因为它不是唯一的。您可以使用 DTO 执行此操作，但 JPA 无法处理投影和join fetch同时，您将只会获得 actor.id 和 meanOfPayment.id ，否则您将进行多项选择。在这个用例中，多选可能不是交易杀手，但您必须自己决定所有这些。当然，您也可以将ActorContentactor.id、meanOfPayment.id 和 createDate 的组合设为主键，这将具有成为上述约束的额外好处。

这些是Entities我一起工作的。

@Entity
public class Actor {
    @Id @GeneratedValue(strategy=GenerationType.IDENTITY)
    private Integer id;

@Entity
public class MeanOfPayment {
    @Id @GeneratedValue(strategy=GenerationType.IDENTITY)
    private Integer id;

@Entity
public class ActorContent {
    @Id @GeneratedValue(strategy=GenerationType.IDENTITY)
    private Integer id;

    @ManyToOne
    private Actor actor;
    @ManyToOne
    private MeanOfPayment meanOfPayment;

    private BigDecimal amount;
    @Temporal(TemporalType.TIMESTAMP)
    private Date createDate;
于 2018-10-24T03:43:34.277 回答