具有聚合属性的 EJB3/JPA 实体

发布于 2024-09-19 20:17:54 字数 817 浏览 1 评论 0原文

我想知道是否有一种方法可以在 One2Many 关系中获得作为多方聚合的一方的字段。

让我们看下面的例子：

@Entity
public class A {
 @Id
 private Long id;
 @OneToMany (mappedBy="parentA")
 private Collection<B> allBs;

 // Here I don't know how to Map the latest B by date
 private B latestB;
    // Acceptable would be to have : private Date latestBDate;
}

@Entity
public class B {
 @Id
 private Long id;
 private Date date;
 @ManyToOne (targetEntity=A.class)
 private A parentA;
}

我的问题是如何在 A 实体对象中映射latestB 字段而不进行任何反规范化（不使字段与触发器/侦听器保持同步）？

也许这个问题给出了一些答案，但实际上我不明白它是如何工作的，因为我仍然希望能够获取所有孩子的物体。

感谢您的阅读/帮助。

PS：我使用hibernate作为ORM/JPA提供者，因此如果不存在JPA解决方案，可以提供Hibernate解决方案。
PS2：或者只是告诉我我不应该这样做（当然有争论）;-)

原文

I wanted to know if there is a way to get in a One2Many relationship a field of the One side that is an aggregate of the Many side.

Let's take the following example:

@Entity
public class A {
 @Id
 private Long id;
 @OneToMany (mappedBy="parentA")
 private Collection<B> allBs;

 // Here I don't know how to Map the latest B by date
 private B latestB;
    // Acceptable would be to have : private Date latestBDate;
}

@Entity
public class B {
 @Id
 private Long id;
 private Date date;
 @ManyToOne (targetEntity=A.class)
 private A parentA;
}

My question is how can I make the mapping of the field latestB in the A entity object without doing any de-normalization (not keeping in sync the field with triggers/listeners)?

Perhaps this question gives some answers, but really I don't understand how it can work since I still want to be able to fetch all childs objects.

Thanks for reading/helping.

PS: I use hibernate as ORM/JPA provider, so an Hibernate solution can be provided if no JPA solution exists.
PS2: Or just tell me that I should not do this (with arguments of course) ;-)

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少钕鈤記 2024-09-26 20:17:54

我使用hibernate作为ORM/JPA提供者，因此如果不存在JPA解决方案，可以提供Hibernate解决方案。

使用@Formula 可以实现可接受的解决方案（即获取最新B 的Date）。

@Entity
public class A {
    @Id
    private Long id;
    @OneToMany (mappedBy="parentA")
    private Collection<B> allBs;

    @Formula("(select max(b.some_date) from B b where b.a_id = id)")
    private Date latestBDate;
}

参考资料

Hibernate 注解参考指南
- 2.4.3.1。公式

资源

Hibernate 派生属性 - 性能和可移植性

I use hibernate as ORM/JPA provider, so an Hibernate solution can be provided if no JPA solution exists.

Implementing the acceptable solution (i.e. fetching a Date for the latest B) would be possible using a @Formula.

@Entity
public class A {
    @Id
    private Long id;
    @OneToMany (mappedBy="parentA")
    private Collection<B> allBs;

    @Formula("(select max(b.some_date) from B b where b.a_id = id)")
    private Date latestBDate;
}