java hibernate：在多态hql查询中选择鉴别器列
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java hibernate: selecting the discriminator column in polymorphic hql query
提问by flybywire
In hibernate, I want to select the discriminator value. Something like
在休眠中，我想选择鉴别器值。就像是
select discriminator, id, name, age from Animal
select discriminator, id, name, age from Animal
The idea is to send the result of this query to the client side, so that I can display a different icon based on the value of the discriminator column (i.e. cat, dog, elephant, etc).
这个想法是将这个查询的结果发送到客户端，这样我就可以根据鉴别器列的值（即猫、狗、大象等）显示不同的图标。
Is that possible? How?
那可能吗？如何？
回答by axtavt
You can do it as follows:
你可以这样做：
select a.class, a.id, a.name, a.age from Animal a
From Hibernate Documentation:
从休眠文档：
The special property class accesses the discriminator value of an instance in the case of polymorphic persistence.
在多态持久化的情况下，特殊属性类访问实例的鉴别器值。
回答by Santiago Taba
Hibernate query objects, does not know columns. So unless you have a property named discriminator in your Animal object you cant do that. You can do the query in sql or get the entire object and then to get the inherited type, for that you can use "instanceof"
Hibernate 查询对象，不知道列。所以除非你的 Animal 对象中有一个名为 discriminator 的属性，否则你不能这样做。您可以在sql中进行查询或获取整个对象然后获取继承的类型，为此您可以使用“instanceof”
