sql - JPA标准API : query property of subclass

Question

我有一个这样的类结构:

@Entity
@Inheritance(strategy = InheritanceType.JOINED)
public abstract class Article {
   private String aBaseProperty;
}

@Entity
public class Book extends Article {
   private String title;
}

@Entity
public class CartItem {
   @ManyToOne(optional = false)
   public Article article;
}

我尝试了以下操作，以接收所有引用了 CartItems的 Book的 title = 'Foo':

CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery<CartItem> query = builder.createQuery(CartItem.class);
Root<CartItem> root = query.from(CartItem.class);
builder.equal(root.get("article").get("title"), "Foo");
List<CartItem> result = em().createQuery(query).getResultList();

但不幸的是，这会导致错误(对我来说很有意义，因为 title在 Book中，而不在 Article中……):

java.lang.IllegalArgumentException: Could not resolve attribute named title    at org.hibernate.ejb.criteria.path.SingularAttributePath.locateAttributeInternal(SingularAttributePath.java:101)    at org.hibernate.ejb.criteria.path.AbstractPathImpl.locateAttribute(AbstractPathImpl.java:216)    at org.hibernate.ejb.criteria.path.AbstractPathImpl.get(AbstractPathImpl.java:189)...

However, I was able to achieve what I want using the following HQL:

SELECT c, a FROM CartItem c INNER JOIN c.article a WHERE a.title = ?

那么后者为何起作用，我是否可以使用Criteria API达到类似的目的？

Answer 1

我遇到了同样的问题，并感谢chris(请参阅JPA Criteria API where subclass)找到了解决方案。

为此，您需要JPA 2.1，并使用 CriteriaBuilder.treat() 方法之一。只需将您的builder.equal...行替换为:

builder.equal(builder.treat(root.get("article"), Book.class).get("title"), "Foo");