有一个数据库表，结构为：（3个字段）
from 从一方 字符型
to 到另一方 字符型
time 时间 时间型
数据：
from to time
111 222 2002-10-10
111 333 2002-10-11
111 444 2002-10-12
111 222 2002-10-13
111 222 2002-10-14
111 333 2002-10-15
222 111 2002-10-16
222 333 2002-10-17
222 555 2002-10-18
444 222 2002-10-19
444 222 2002-10-20
666 222 2002-10-21
要查询from与to字段合并后出现次数在n次以上的结果集
如：
n=2结果集为：
from to time
111 222 2002-10-10
111 333 2002-10-11
111 222 2002-10-13
111 222 2002-10-14
111 333 2002-10-15
444 222 2002-10-19
444 222 2002-10-20
n=3结果集为：
from to time
111 222 2002-10-10
111 222 2002-10-13
111 222 2002-10-14
不知这种情况的SQL语句如何写？
---------------------------------------------------------------
declare @n int
set @n=2 -- change 2 to 3 for the second case
select * from #mytable t1
where
(select count(*)
from #mytable t2
where t2.[from]=t1.[from] and t2.[to]=t1.[to]) >= @n
---------------------------------------------------------------
select * from 你的表 b where exists (select 1 from (select [from],[to] from 你的表 group by [from],[to] having count(*)>=@你的数) a where a.[from]=b.[from] and a.to=b.to)