public List getAllMess(final Party party) {
		List list_temp = getHibernateTemplate().executeFind(new HibernateCallback() {
			public Object doInHibernate(Session session)
					throws HibernateException, SQLException {

				Query query = session
						.createQuery("select m.messMsg as name,count(m.messMsg) as count from ReceivedMessage m where m.party.id=:party group by m.messMsg");
				// Query query = session.createQuery("from ReceivedMessage where party_id=:party");
				query.setLong("party", party.getId());
				
				return query.list();
			}

		});
           return list_temp;
	}

select m.messMsg as name,count(m.messMsg) as count from ReceivedMessage m where m.party.id=:party group by m.messMsg

//这里用的是group by

ReceivedMessage 里面的messMsg有N多相同的,这样地得到的就是重复出现的数量和列值
my god,一上午的劳动成果啊,开始觉得用distinct,结果查了N久也不出,最后看了抛出异常的爱的写法,终于写出来了,在这里感谢他的帖子
评论
不是流氓 2008-06-23
对啊,我开始也是查distinct,我日,那个不总是搞不定,就 group by了
spiritfrog 2008-06-23
果然是用group by,lz应该写清楚用了什么方法。
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