JPA 2 无显式选择且隐式冷无法确定

Question

我正在尝试获取某个文件夹中的所有用户，该用户是在特定日期之后创建的。用户和文件夹之间的关系位于单独的表中。

这是我提出的查询，但它引发了异常

无显式选择且隐式冷端未确定

代码

@Override
public List<RetailPostUserTbl> getNewUsersForSiteSince( Date date, Integer siteId )
{
    List<RetailPostUserTbl> toReturn = new ArrayList<RetailPostUserTbl>();
    EntityManager em = getEntityManager();
    CriteriaBuilder cb = em.getCriteriaBuilder();

    Class<RpUserFolderMapTbl> userFolderPC = userFolderMapDAO.getPersistentClass();

    CriteriaQuery<RpUserFolderMapTbl> mapQuery = cb.createQuery( userFolderPC );
    Root<RpUserFolderMapTbl> root = mapQuery.from( userFolderPC );
    Path<Integer> folderIdPath = root.get( RpUserFolderMapTbl_.folder ).get( FolderTbl_.folderId );

    Predicate folderCondition = cb.equal( folderIdPath, siteId );

    Subquery<RetailPostUserTbl> rpSubQ = mapQuery.subquery( persistentClass );
    Root<RetailPostUserTbl> subQRoot = rpSubQ.from( persistentClass );
    Path<UserTbl> userPath = subQRoot.get( RetailPostUserTbl_.user );
    Path<Date> userCreatedPath = userPath.get( UserTbl_.userCreateDate );
    Predicate userCreateDateCondition = cb.greaterThanOrEqualTo( userCreatedPath, date );
    rpSubQ.where( userCreateDateCondition );

    mapQuery.where( cb.and( folderCondition, cb.exists( rpSubQ ) ) );

    TypedQuery<RpUserFolderMapTbl> query = em.createQuery( mapQuery );
    List<RpUserFolderMapTbl> results = query.getResultList();
    for ( RpUserFolderMapTbl result : results )
    {
        RetailPostUserTbl rpuser = result.getUser().getRetailPostUser();
        toReturn.add( rpuser );
    }
    return toReturn;
}

有人知道为什么这不起作用吗？

Answer 1

您还应该为“子查询”明确设置选择。

rpSubQ.select(subQRoot);

Answer 2

我今天遇到了完全相同的错误。有趣的是，我从 Hibernate 3.6.3.Final 文档中获取了示例。他们的例子是：

CriteriaQuery query = builder.createQuery();
Root<Person> men = query.from( Person.class );
Root<Person> women = query.from( Person.class );
Predicate menRestriction = builder.and(
    builder.equal( men.get( Person_.gender ), Gender.MALE ),
    builder.equal( men.get( Person_.relationshipStatus ), RelationshipStatus.SINGLE )
);
Predicate womenRestriction = builder.and(
    builder.equal( women.get( Person_.gender ), Gender.FEMALE ),
    builder.equal( women.get( Person_.relationshipStatus ), RelationshipStatus.SINGLE )
);
query.where( builder.and( menRestriction, womenRestriction ) );

我为“修复”错误所做的就是显式选择根。注意我必须创建一个根来解决这个问题。这是我的例子：

CriteriaQuery query = builder.createQuery();
Root<Person> personRoot = query.from( Person.class );
Predicate menRestriction = builder.and(
    builder.equal( personRoot.get( Person_.gender ), Gender.MALE ),
    builder.equal( personRoot.get( Person_.relationshipStatus ), RelationshipStatus.SINGLE )
);
Predicate womenRestriction = builder.and(
    builder.equal( personRoot.get( Person_.gender ), Gender.FEMALE ),
    builder.equal( personRoot.get( Person_.relationshipStatus ), RelationshipStatus.SINGLE )
);
query.select(personRoot);
query.where( builder.and( menRestriction, womenRestriction ) );

我不明白为什么不能进行隐式选择。在 Hibernate 的示例中，唯一使用的类是 Person.class。当我进一步挖掘时，我会更新我的回复。

Answer 3

查询 roots 指定其上的域对象查询被评估。查询根是 Root 接口的实例，必须显式声明，因此

CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<T> criteriaQuery = criteriaBuilder.createQuery(type);
list.forEach(i -> entityManager
         .createQuery(criteriaQuery.where(
              criteriaBuilder.equal(criteriaQuery.from(type).get(field), i))).getResultList());

您必须将 root 显式定义为

CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<T> criteriaQuery = criteriaBuilder.createQuery(type);
Root<T> root = criteriaQuery.from(type);
list.forEach(i -> entityManager
        .createQuery(criteriaQuery.where(
            criteriaBuilder.equal(root.get(field), i))).getResultList());