Question

我正在尝试使用JPA Criteria API实现类似以下内容：

SELECT b FROM Box b JOIN SpecialItem s WHERE s.specialAttr = :specialAttr

对象是

箱

@Entity
public class Box implements Serializable {
  ...
  @ManyToOne
  @JoinColumn( name = "item_id" )
  Item item;
  ...
}

物品

@Entity
@Inheritance( strategy = InheritanceType.JOINED )
public class Item implements Serializable {
  @Id
  private String id;
  ...
}

SpecialItem

@Entity
public class SpecialItem extends Item {
  private String specialAttr;
  ...
}

我的尝试

EntityManager em = getEntityManager();
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery cq = cb.createQuery( Box.class );
Root from = cq.from( Box.class );

// Nothing to specify SpecialItem over Item!      
Join join = from.join("item", JoinType.LEFT);

// java.lang.IllegalArgumentException: Unable to 
// resolve attribute [specialAttr] against path [null]      
Path p = join.get( "specialAttr" );

Predicate predicate = cb.equal( p, "specialValue" );
cq.where( predicate );

因为specialAttr不是类Item的成员，所以抛出异常并不奇怪。

如何返回包含Box的所有SpecialItem，其中SpecialItem.specialAttr具有某些价值？

Answer 1

如果使用JPA 2.1，您可以使用

"SELECT b FROM Box b WHERE TREAT(b.item as SpecialItem).specialAttr = :specialAttr"

或

CriteriaQuery<Box> q = cb.createQuery(Box.class);
Root<Box> box= q.from(Box.class);
Join<Box, Item > order = box.join("item");
q.where(cb.equal(cb.treat(order, SpecialItem.class).get("specialAttr"),
    qb.parameter(String.class, "specialAttr")));
q.select(Box);

Answer 2

我只想通过生成的元模型扩展Chris对Criteria API的答案。

CriteriaQuery<Box> q = cb.createQuery(Box.class);
Root<Box> box= q.from(Box.class);
Join<Box, Item> order = box.join(Box_.item);
q.where(cb.equal(cb.treat(order, SpecialItem.class).get(SpecialItem_.specialAttr), "searchValue");
q.select(Box);

如何在CriteriaQuery中引用查询超类的子类特定字段？

2 个答案: