select top 6 with ties * from mem order by [money] desc
--假設ID自增 select top 6 * from mem t where not exists(select 1 from mem where [money]=t.[money] and id>t.id) order by [money] desc
select top 6 *from men order by money desc是这样吗
select top 6 * from mem t where id=(select max(id) from mem where [money]=t.[money]) order by [money] desc
--我的方法很笨,楼主可以参考一下 --假如楼主你的表是user_info --只要从user_info中找出虚拟币为(最高的6个虚拟币中的其中一个)的客户即可 select * from user_info where money in (select top 6 money from user_info order by money desc )--最活跃的6个人的虚拟币
--没看到楼主的表名 --换了 select * from mem where money in (select top 6 money from mem order by money desc )--最活跃的6个人的虚拟币
from mem
order by [money] desc
select top 6 *
from mem t
where not exists(select 1 from mem where [money]=t.[money] and id>t.id)
order by [money] desc
order by money desc是这样吗
top 6 *
from
mem t
where
id=(select max(id) from mem where [money]=t.[money])
order by
[money] desc
--我的方法很笨,楼主可以参考一下
--假如楼主你的表是user_info
--只要从user_info中找出虚拟币为(最高的6个虚拟币中的其中一个)的客户即可
select * from user_info
where money in (select top 6 money
from user_info order by money desc
)--最活跃的6个人的虚拟币
--没看到楼主的表名
--换了
select * from mem
where money in (select top 6 money
from mem
order by money desc
)--最活跃的6个人的虚拟币