例:

1、select subs(xh,1,4) as bmd,count(xh) as rs from ls group by bmd   &&&能通过
2、select subs(xh,1,4) as bmd,count(xh) as rs from ls group by 1     &&也能通过
3、select subs(xh,1,4) as bmd,count(xh) as rs from ls group by subs(xh,1,4) &&不能通过为什么？

3、select subs(xh,1,4) as bmd,count(xh) as rs from ls group by bmd
改成这样试试

这当然能通过，我已有例子。

这个不难理解，group by是对形成的查询表进行分组，在新的查询中就没有xh这个字段，当然不行，如果改成
3、select subs(xh,1,4) as bmd,count(xh) as rs from ls group by subs(bmd,1,4) 应该可以

通不过，提示找不到列'        '

那就用1、2

SELECT [ALL | DISTINCT] [TOP nExpr [PERCENT]] Select_List_Item [, ...]
   FROM [FORCE] Table_List_Item [, ...]
      [[JoinType] JOIN DatabaseName!]Table [[AS] Local_Alias]
      [ON JoinCondition [AND | OR [JoinCondition | FilterCondition] ...]
   [WITH (BUFFERING = lExpr)]
   [WHERE JoinCondition | FilterCondition [AND | OR JoinCondition | FilterCondition] ...]
   [GROUP BY Column_List_Item [, ...]] [HAVING FilterCondition [AND | OR ...]]
   [UNION [ALL] SELECTCommand]
   [ORDER BY Order_Item [ASC | DESC] [, ...]]
   [INTO StorageDestination | TO DisplayDestination]
   [PREFERENCE PreferenceName] [NOCONSOLE] [PLAIN] [NOWAIT]

人家说得很明白，是Column_list_item,不是Expression。分组字段可以这样存在：
sele AA+BB,max(CC) from alias() group by AA,它要么是查询结果中的一个字段，要么就是隐含在查询结果中。

顶楼的第3条代码犯了语法错误，7楼已经解释得比较明了。

以前我曾经遇到过这个问题，没有弄懂，今天刚开始没有仔细看。现在总算弄明白了

'隐含在查询结果中'是何种样子？请明示。
AA+BB应是二个字段组合后的结果，哪其中AA是隐含在查询结果中还是查询结果中的一个字段？


[ 本帖最后由 sylknb 于 2013-8-15 13:52 编辑 ]