如何在Spring JpaRepository中使用JPQL选择组中的最新记录？

我认为在关系代数方面，您要ActorContent 减去的集合是ActorContent受actor = actor和meanOfPayment = meanOfPayment和createDate <createDate约束的集合。因此，思考的方法是从ActorContentwith的叉积获得第二组ac1.meanOfPayment = ac2.meanOfPayment and ac1.actor = ac2.actor and ac1.createDate < ac2.createDate。然后从中减去该集合ActorContent。我没有看过它是否比使用MAX和Group By例如更有效：

@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，分别代表第一个演员和他的唯一meanOfPayment，第二个演员和他对所有三个meanOfPayments的最新付款。

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，您可能希望通过联接获取Actor和MeanOfPayment实例来优化查询。通过示例：

@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，如果您想要特定的内容，Actor则只需将其添加到where子句中即可。

@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组合使用相同的createDate时遇到问题，则可以采用几种不同的方法来处理。首先，如果您具有逻辑约束以至于不想处理那些重复项，那么您可能还应该具有数据库约束，这样就不会得到它们并确保您首先不要创建它们。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一个ActorContent非重复，但是您必须省略id字段，因为它不是唯一的。您可以使用DTO进行此操作，但JPA无法处理投影和join fetch同时，因此您只会获得actor.id和meanOfPayment.id，否则您将进行多项选择。在此用例中，多重选择可能不是交易杀手，但您必须自己决定所有这一切。当然，您也可以将ActorContentactor.id，meanOfPayment.id和createDate组合在一起作为主键，这将具有上述约束的好处。

这些是Entities我合作过的。

@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;