SQL传入时间获取到时间的周一和周日
declare @time datetime declare @timeMonday datetime set @time='2013-11-07' select @timeMonday=dateadd(day,1 - (datepart(weekday,@time) + @@datefirst - 2) % 7 - 1,@time) select @timeMonday --周一的时间 select dateadd(day,6,convert(varchar(20), @timeMonday, 23)) --周日的时间