如何指定 JPA 标准 API 和 "MEMBER IN" 以及映射超类中的多对多关系？

Question

我想查询 ConcreteEntity where AUser user IS MEMBER of ConcreteEntity.createdBy with

@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();

我无法从 SetAttribute 过渡到 PluralAttribute。我知道我不应该 post 质疑无法编译的代码，但尽管我会说我对泛型有一定的了解，但我在这里不明白。

我正在使用 Eclipselink 2.5.2 和 JPA 2.1。

Answer 1

查看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);

第二个应该是：

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

所以，恕我直言，这不是你的错，API 有问题。

需要解决方法。只需擦除编译时类型（它在运行时被擦除，无论如何）。

你必须做的：

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");

如何指定 JPA 标准 API 和 "MEMBER IN" 以及映射超类中的多对多关系？

How specify a JPA criteria API with "MEMBER IN" and a many-to-many relationship in a mapped superclass?

jpa

criteria

eclipselink

criteria-api