JPA Criteria使用单表继承查询实体层次结构 [英] JPA Criteria Query over an entity hierarchy using single table inheritance

问题描述

假设我有以下实体：

@Entity
@Inheritance(strategy = SINGLE_TABLE)
@DiscriminatorColumn(name = "type")
public abstract class BaseEntity { 
    private Date someDate;
    private Date otherDate;
    private boolean flag;
}

@Entity
@DiscriminatorValue("entity1")
public class Entity1 extends BaseEntity { 
    private String someProperty;
}

@Entity
@DiscriminatorValue("entity2")
public class Entity2 extends BaseEntity { 
    private String otherProperty;
}

我正在尝试构建一个基于以下方式返回BaseEntity实例的条件查询BaseEntity中的属性和两个子类。所以基本上我正在寻找对应于这个伪SQL的条件查询：

I'm trying to build a criteria query that returns instances of BaseEntity based on properties in BaseEntity and both of the subclasses. So essentially I'm looking for a criteria query that corresponds to this pseudo-SQL:

SELECT * FROM <BaseEntity table name>
WHERE someDate < ? AND otherDate > ? AND flag = ?
AND someProperty = ? AND otherProperty = ?;

我宁愿不构建两个单独的查询，因为它们有很多重叠（即大多数属性）在基类）。但是，如果我将BaseEntity声明为root，我还没有想出一种在查询中引用子类属性的方法。是否可以建立这样的标准查询？

I'd rather not build two separate queries since they have so much overlap (i.e. most of the properties are in the base class). However, I haven't figured out a way to reference the subclass properties in the query if I declare BaseEntity as the root. Is it possible to build a criteria query like this?

更新：

也许一些代码会澄清这个问题。我基本上喜欢这样做：

Maybe some code would clarify the question. I'd essentially like to do something like this:

CriteriaBuilder builder = ...;
CriteriaQuery<BaseEntity> query = ...;
Root<BaseEntity> root = ...;

query.select(root).where(builder.and(
        builder.lessThan(root.get(BaseEntity_.someDate), new Date()),
        builder.greaterThan(root.get(BaseEntity_.otherDate), new Date()),
        builder.isTrue(root.get(BaseEntity_.flag)),
        builder.equal(root.get(Entity1_.someProperty), "foo"),   <-- This won't work
        builder.equal(root.get(Entity2_.otherProperty), "bar")   <-- Neither will this
));

现在，我理解为什么上面的代码示例不起作用，但我想知道如果有办法解决它。

Now, I understand why the above code sample doesn't work, but I'd like to know if there's a way to get around it.

推荐答案

我设法通过将BaseEntity根目录转换为对应的新根来解决这个问题。 CriteriaBuilder.treat（），如下所示：

I managed to solve this by downcasting the BaseEntity root into new roots that correspond to the subclass types with CriteriaBuilder.treat() like this:

CriteriaBuilder builder = ...;
CriteriaQuery<BaseEntity> query = ...;
Root<BaseEntity> root = ...;
Root<Entity1> entity1 = builder.treat(root, Entity1.class);
Root<Entity2> entity2 = builder.treat(root, Entity2.class);

query.select(root).where(builder.and(
        builder.lessThan(root.get(BaseEntity_.someDate), new Date()),
        builder.greaterThan(root.get(BaseEntity_.otherDate), new Date()),
        builder.isTrue(root.get(BaseEntity_.flag)),
        builder.equal(entity1.get(Entity1_.someProperty), "foo"),
        builder.equal(entity2.get(Entity2_.otherProperty), "bar")
));

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