取与指定时间最近的数据集

if object_id('[tabletest]') is not null drop table [tbl]

create table [tabletest]([DATime] datetime,[DaWeight] varchar(5))

insert [tabletest]

select '2012-03-15 14:24:00.000','300Kg' union all

select '2012-03-15 14:25:00.000','400Kg'

go

declare  @datimenow  DateTime

set @datimenow=getdate()

select  *   from   tabletest

--直接取最新的记录

select top 1 * from tabletest order by DATime desc

--取小于当前时间的最新的记录

select top 1 * from tabletest where DATime < @datimenow order by DATime desc

--如果是取离当前时间最近的,包括大于当前时间,按照秒来算.

--如果只有一条记录.

select top 1 * from tabletest order by abs(datediff(ss,datime,@datimenow)) 

--如果可能存在多条记录.

select * from tabletest where abs(datediff(ss,datime,@datimenow)) = (select min(abs(datediff(ss,datime,@datimenow))) from tabletest) 

--根据差值排序,取序列号。根据指定序列号返回数据。

select * from(

select ROW_NUMBER()over(order by datediff(mi,datime,getdate())) as id,

* from tabletest) a where id=1