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