16

我正在尝试获取在某个日期之后创建用户的文件夹的所有用户。用户和文件夹之间的关系存在于单独的表中。

这是我提出的查询，但它引发了异常

没有明确的选择和隐含的一冷未定

编码

@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;
}

任何人都知道为什么这不起作用？

4

2 回答 2

30

您还应该为“子查询”设置显式选择。

rpSubQ.select(subQRoot);
于 2014-09-16T08:46:58.183 回答
8

我今天有完全相同的错误。有趣的是，我从 Hibernate 3.6.3.Final 文档中获取了我的示例。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”错误所做的是明确选择根。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 ) );

我无法弄清楚为什么无法进行隐式选择。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。

于 2011-04-19T21:58:53.333 回答