如何使用“MEMBER IN”指定JPA标准API并在映射的超类中指定多对多关系？

Question

我想查询的ConcreteEntity.createdByConcreteEntity其中AUser userIS MEMBER与

@Entity 
public class AUser implements Serializable { 
    @Id 
    @GeneratedValue 
    private Long id; 
    private String property; 

和

@MappedSuperclass 
public class AbstractEntity implements Serializable { 
    private static final long serialVersionUID = 1L; 
    @Id 
    @GeneratedValue 
    private Long id; 
    @ManyToMany 
    private Set<AUser> createdBy = new HashSet<>(); 

和

@Entity 
@Inheritance(strategy = InheritanceType.SINGLE_TABLE) 
ppublc class ConcreteEntity extends AbstractEntity { 
    private String property1; 

即

EntityManager em = ...; //set to null or else 
AUser user = ...; //set to null or else 
CriteriaBuilder cb = em.getCriteriaBuilder(); 
CriteriaQuery<ConcreteEntity> criteria = cb.createQuery(ConcreteEntity.class); 
Root<ConcreteEntity> concreteRoot = criteria.from(ConcreteEntity.class); 
criteria.select(concreteRoot); 
SetAttribute<AbstractEntity, AUser> setAttribute = ConcreteEntity_.createdBy; 
PluralAttribute<ConcreteEntity, Set<AUser>, AUser> pluralAttribute = setAttribute; 
    //richtercloud/jpa/criteria/api/metamodel/member/of/NewMain.java:[40,77] error: incompatible types: SetAttribute<AbstractEntity,AUser> cannot be converted to PluralAttribute<ConcreteEntity,Set<AUser>,AUser> 
Expression<Set<AUser>> createdByExpression = concreteRoot.get(pluralAttribute); 
criteria.where(cb.isNotMember(user, createdByExpression)); 
TypedQuery<ConcreteEntity> query = em.createQuery(criteria); 
List<ConcreteEntity> result = query.getResultList(); 

我没有从SetAttribute转换到PluralAttribute。我知道我不应该用无法编译的代码发布问题，但尽管我会说我对泛型有一个公正的理解，但我不明白。我在https://github.com/krichter722/jpa-criteria-api-metamodel-member-of提供了一个示例项目。

我正在使用Eclipselink 2.5.2和JPA 2.1。

Answer 1

看看Path.get方法：

<Y> Path<Y> get(SingularAttribute<? super X, Y> attribute); 

<E, C extends Collection<E>> Expression<C> get(PluralAttribute<X, C, E> collection); 

第二应改为：

<E, C extends Collection<E>> Expression<C> get(PluralAttribute<? super X, C, E> collection); 

因此，恕我直言，这不是你的错，API被窃听。

需要解决方法。只需擦除编译时类型（无论如何它在运行时擦除）。

你要做的：

Expression<Set<AUser>> createdByExpression = ((Root<AbstractEntity>) (Object) root).get(AbstractEntity_.createdBy); 

或

Expression<Set<AUser>> createdByExpression = root.get((SetAttribute<ConcreteEntity, AUser>) (Object) AbstractEntity_.createdBy); 

否则

Expression<Set<AUser>> createdByExpression = root.get("createdBy");