Far*_*san 14 criteria-api jpa-2.0
我正在尝试获取在特定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,但是这个例外就是一个例子
没有明确的选择和隐含的感冒不被确定
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;
}
Run Code Online (Sandbox Code Playgroud)
任何人都知道为什么这不起作用?
我今天遇到了完全相同的错误.有趣的是,我从Hibernate 3.6.3.Final docs中抓住了我的例子.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 ) );
Run Code Online (Sandbox Code Playgroud)
我所做的"修复"错误是明确选择根.注意我必须创建一个root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 ) );
Run Code Online (Sandbox Code Playgroud)
我无法弄清楚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.
| 归档时间: |
|
| 查看次数: |
13069 次 |
| 最近记录: |