如何使用 JPA 和 Hibernate 映射计算属性 [英] How to map calculated properties with JPA and Hibernate

问题描述

我的 Java bean 有一个 childCount 属性.此属性未映射到数据库列.相反，它应该由数据库使用 COUNT() 函数计算，该函数对我的 Java bean 及其子级的连接进行操作.如果可以按需/懒惰"地计算此属性会更好，但这不是强制性的.

My Java bean has a childCount property. This property is not mapped to a database column. Instead, it should be calculated by the database with a COUNT() function operating on the join of my Java bean and its children. It would be even better if this property could be calculated on demand / "lazily", but this is not mandatory.

在最坏的情况下，我可以使用 HQL 或 Criteria API 设置此 bean 的属性，但我不想这样做.

In the worst case scenario, I can set this bean's property with HQL or the Criteria API, but I would prefer not to.

Hibernate @Formula 注释可能会有所帮助，但我几乎找不到任何文档.

The Hibernate @Formula annotation may help, but I could barely find any documentation.

非常感谢任何帮助.谢谢.

Any help greatly appreciated. Thanks.

推荐答案

JPA 不提供对派生属性的任何支持，因此您必须使用提供者特定的扩展.正如您所提到的，@Formula 在使用 Hibernate 时非常适合于此.您可以使用 SQL 片段:

JPA doesn't offer any support for derived property so you'll have to use a provider specific extension. As you mentioned, @Formula is perfect for this when using Hibernate. You can use an SQL fragment:

@Formula("PRICE*1.155")
private float finalPrice;

甚至是对其他表的复杂查询:

Or even complex queries on other tables:

@Formula("(select min(o.creation_date) from Orders o where o.customer_id = id)")
private Date firstOrderDate;

其中id是当前实体的id.

以下博客文章值得一读:Hibernate 派生属性 - 性能和便携性.

The following blog post is worth the read: Hibernate Derived Properties - Performance and Portability.

没有更多细节，我无法给出更准确的答案，但上面的链接应该会有所帮助.

Without more details, I can't give a more precise answer but the above link should be helpful.

• 第 5.1.22.列和公式元素(Hibernate Core 文档)
• 第 2.4.3.1.公式(Hibernate Annotations 文档)

• Section 5.1.22. Column and formula elements (Hibernate Core documentation)
• Section 2.4.3.1. Formula (Hibernate Annotations documentation)

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