Java /JPA | 具有指定继承类型的查询
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Java /JPA | Query with specified inherited type
提问by redbull
I am building a query on a generic table "Sample" and I have several types which inherit from this table "SampleOne", "SampleTwo". I require a query like :
我正在一个通用表“Sample”上构建一个查询，我有几种类型从这个表“SampleOne”、“SampleTwo”继承而来。我需要一个查询，如：
select s from Sample where s.type = :type
where type would be a discriminator value of the table. Is it possible in any way ( and avoid to create an entity specific queries, one for each SampleOne, SampleTwo... etc )
其中 type 将是表的鉴别器值。是否有可能以任何方式（并避免创建实体特定的查询，每个 SampleOne、SampleTwo... 等）
I would greatly appreciate any input in this topic,
我将不胜感激对此主题的任何投入，
Kind regards, P.
亲切的问候，P。
回答by axtavt
In JPA 2.0 you can use TYPEexpression (though currently it doesn't work with parameters in Hibernate, see HHH-5282):
在 JPA 2.0 中，您可以使用TYPE表达式（尽管目前它不适用于 Hibernate 中的参数，请参阅HHH-5282）：
select s from Sample s where TYPE(s) = :type
The similar Hibernate-specific expression is .class:
类似的 Hibernate 特定表达式是.class：
select s from Sample s where s.class = :type
回答by Sean Patrick Floyd
Here's the relevant section of the Java EE 6 tutorial:
这是Java EE 6 教程的相关部分：
Abstract Entities
An abstract class may be declared an entity by decorating the class with
@Entity. Abstract entities are like concrete entities but cannot be instantiated.Abstract entities can be queried just like concrete entities. If an abstract entity is the target of a query, the query operates on all the concrete subclasses of the abstract entity:
抽象实体
抽象类可以通过用 装饰类来声明为实体
@Entity。抽象实体就像具体实体，但不能被实例化。抽象实体可以像具体实体一样被查询。如果抽象实体是查询的目标，则查询对抽象实体的所有具体子类进行操作：
@Entity
public abstract class Employee {
@Id
protected Integer employeeId;
...
}
@Entity
public class FullTimeEmployee extends Employee {
protected Integer salary;
...
}
@Entity
public class PartTimeEmployee extends Employee {
protected Float hourlyWage;
}
If I read this right, your query:
如果我没看错，您的查询：
select s from Sample where s.type = :type
Should only return elements of the specified subtype if typeis the discriminator column, so the only thing that's left for you to do is to cast the result list to your requested sub type.
如果type是鉴别器列，则只应返回指定子类型的元素，因此您唯一要做的就是将结果列表转换为您请求的子类型。
回答by Wim De Rammelaere
You still have to be carefull in Hibernate 4.3.7, because there is still an issue with the implementation of TYPE(), for example:
在 Hibernate 4.3.7 中你仍然要小心，因为在实现上仍然存在问题TYPE()，例如：
from SpoForeignPilot sfp where TYPE(sfp.partDocument) = :type
This query doesn't work as it incorrectly checks the type of SpoForeignPilotand not the type of the document.
此查询不起作用，因为它错误地检查SpoForeignPilot了文档的类型而不是文档的类型。
You can workaround this issue by doing something like this:
您可以通过执行以下操作来解决此问题：
select sfp from SpoForeignPilot sfp join sfp.partDocument doc where TYPE(doc) = :type
回答by Gourav Jindal
Do this in your repository
在您的存储库中执行此操作
@Query("SELECT p FROM BaseUserEntity p where p.class=:discriminatorValue")
public List<BaseUserEntity> findByDiscriminatorValue(@Param("discriminatorValue") String discriminatorValue);
Where BaseUserEntityis your parent entity
BaseUserEntity你的父实体在哪里
