如何在JPA 2.0中创建'类似'的查询实例？ [英] How to create an 'instance of' -like query in JPA 2.0?

问题描述

假设我们有一个抽象的@Entity Animal，以及几个扩展Animal的实体类，包括Dog，Cat，Monkey和Bat。

Say we've got an abstract @Entity Animal, and several entity classes that extend Animal, including Dog, Cat, Monkey and Bat.

我如何过滤结果基于扩展实体的类？

How can I filter the results based on the extending entity's class?

示例：
有复选框，用户可以选择要检索的实体。

Example: There are checkboxes where the user can select which entities to retrieve.

[ ] Dog
[X] Cat
[X] Monkey
[ ] Bat

现在我想检索动物类。我可以在查询中放入什么样的查询参数，以便只返回Cat和Monkey对象？

Now I want to retrieve the entities with a (Named)Query defined in the Animal class. What kind of query parameters can I put into the query so that only the Cat and Monkey objects will be returned?

推荐答案

我'我不完全确定它是由JPA支持的，但是在Hibernate中这样做的方式，无论继承策略如何，因此即使你没有鉴别器（或者没有将它作为属性映射）也是使用隐式类属性：

I'm not absolutely sure it's supported by JPA, but the way to do it in Hibernate, regardless of the inheritance strategy, and thus even if you don't have a discriminator (or didn't map it as a property) is to use the implicit class property :

String jpql = "select a from Animal a where a.class in (:classes)";
Query q = em.createQuery(jpql).setParameter("classes", 
                                            Arrays.asList(Cat.class, Monkey.class));

编辑：

我刚刚在JPA2中发现使用TYPE运算符：

I just found it's possible in JPA2 using the TYPE operator :

String jpql = "SELECT a FROM Animal a WHERE TYPE(a) IN :classes";
Query q = em.createQuery(jpql).setParameter("classes", 
                                            Arrays.asList(Cat.class, Monkey.class));

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