Hibernate HQL按多对多集合属性排序

我有一个名为

User

的hibernate对象，它与另一个名为

Post

的hibernate对象有多对多关系。这种关系称为

likedPosts

（从

用户的角度）

我想让这个列表按
Post
对象（特别是Post id）中的属性排序。我（天真的）尝试是这样的：

"select user.likedPosts from User user where user.id=:uid 
        order by user.likedPosts.id desc"

但这不起作用-我得到一个例外，告诉我我不允许这样做。我该怎么做呢？谢谢

错误的堆栈跟踪：

org.hibernate.QueryException: illegal attempt to dereference collection [user0_.ID.likedPosts] with element property reference [id] [select user.likedPosts from com.pashash.domain.User user where user.id=:uid order by user.likedPosts.id desc]
    at org.hibernate.hql.ast.tree.DotNode$1.buildIllegalCollectionDereferenceException(DotNode.java:46)
    at org.hibernate.hql.ast.tree.DotNode.checkLhsIsNotCollection(DotNode.java:513)
    at org.hibernate.hql.ast.tree.DotNode.resolve(DotNode.java:221)
    at org.hibernate.hql.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:94)
    at org.hibernate.hql.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:90)
    at org.hibernate.hql.ast.HqlSqlWalker.resolve(HqlSqlWalker.java:728)
    ...

要做到这一点，您需要显式联接：

select post from User user
inner join user.likedPosts post
where user.id = :userId
order by post.id desc

打印excpetion的stacktrace。为什么不让用户使用会话。获取并查看他的喜好（并在内存中排序）？这将是更多的OO风格，等等。。。当用户和他喜欢的人已经在内存中时，速度可能会更快。问得好。两个原因：首先，我希望能够限制从数据库返回的like的数量（这是
createQuery
允许我做的）。第二，这个查询的结果将由hibernate缓存在二级缓存中，因此，我猜在每次请求时使用它比使用它更有效。谢谢，我希望避免使用显式连接，但这完美地解决了我的问题！答案中的输入错误：应该是
where user.id=：userId
org.hibernate.QueryException: illegal attempt to dereference collection [user0_.ID.likedPosts] with element property reference [id] [select user.likedPosts from com.pashash.domain.User user where user.id=:uid order by user.likedPosts.id desc]
    at org.hibernate.hql.ast.tree.DotNode$1.buildIllegalCollectionDereferenceException(DotNode.java:46)
    at org.hibernate.hql.ast.tree.DotNode.checkLhsIsNotCollection(DotNode.java:513)
    at org.hibernate.hql.ast.tree.DotNode.resolve(DotNode.java:221)
    at org.hibernate.hql.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:94)
    at org.hibernate.hql.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:90)
    at org.hibernate.hql.ast.HqlSqlWalker.resolve(HqlSqlWalker.java:728)
    ...

select post from User user
inner join user.likedPosts post
where user.id = :userId
order by post.id desc