Question

从祖先获取对象有什么变化吗？我有这些实体：

@Entity
@Table(name = "SPORT")
@DiscriminatorColumn(name = "sport_type", discriminatorType = DiscriminatorType.STRING, length = 32)
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
public class Sport implements Serializable {

@Column(name = "sport_type", insertable = false, updatable = false)
    private String sportType;

    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.REFRESH)
    @JoinColumn(name = "HEAD_ID", referencedColumnName = "IDENT")
    @ToStringExclude
    protected Headquaters headquaters;

}




@Entity
@DiscriminatorValue(Constants.COST_REPORT_CODE)
public class HandSport extends Sport {

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "COST_TRADE_ID", referencedColumnName = "TRADE_ID")
    private Playground Playground;


}

从这些实体

我已经生成了这个元模型

@Generated(value = "org.hibernate.jpamodelgen.JPAMetaModelEntityProcessor")
@StaticMetamodel(HandSport.class)
public abstract class HandSport_ extends com.test.entity.Sport_ {

    public static volatile SingularAttribute<HandSport, Playground> playground;
}

@Generated(value = "org.hibernate.jpamodelgen.JPAMetaModelEntityProcessor")
@StaticMetamodel(Sport.class)
public abstract class Sport_ {

    public static volatile SingularAttribute<Sport, Headquaters> headquaters;   
    public static volatile SingularAttribute<Sport, String> id;


}

现在我只能拿到游乐场：

EntityGraph<HandSport> fetchGraph = entityManager.createEntityGraph(HandSport.class);
fetchGraph.addSubgraph(HandSport_.playground);
select.distinct(true);
TypedQuery<HandSport> typedQuery = entityManager.createQuery(select)
.setHint("javax.persistence.loadgraph", fetchGraph);

为什么我无法添加

fetchGraph.addSubgraph(Sport_.headquters);

如果我这样做，我可以看到编译错误：

"Cannot resolve method 'addSubgraph(javax.persistence.metamodel.SingularAttribute<com.test.entity.Sport, com.test.entity.Headquateres>)'"

从祖先那里获取值有什么变化吗？或者是否可以创建第二个FetchGraph并通过提示注册两个图形？

Answer 1

JPA API在EntityGraph类的各种方法中使用了泛型。这是reported to the JPA "expert group"早在2015年，但他们并没有费心去解决它。

DataNucleus JPA发布了他们自己的JPA API jar，他们不久前为他们的用户做了修复，但是既然你正在使用Hibernate，那么这对你没有好处。

Criteria api从祖先获取对象

1 个答案: