表結構如下
number date
8 2009/1/11 2:00
7 2009/1/11 5:00
6 2009/1/11 12:00
5 2009/1/11 18:00
4 2009/1/12 4:00
3 2009/1/12 10:00
2 2009/1/12 12:00
1 2009/1/11 17:00
想得到當天的最早時間與最晚時間的number的差值， 即如下的結果:
差
2
3
復制代碼 代碼如下:
create table #date
(
number int identity(1,1) primary key,
date datetime
)
insert into #date select '2009/1/11 17:00'
insert into #date select '2009/1/12 12:00'
insert into #date select '2009/1/12 10:00'
insert into #date select '2009/1/12 4:00'
insert into #date select '2009/1/11 18:00'
insert into #date select '2009/1/11 12:00'
insert into #date select '2009/1/11 5:00'
insert into #date select '2009/1/11 2:00'
select (d2.number-d1.number) number
from
(
select number,date from #date where date in
(select max(date) from #date group by convert(varchar(10),date,120) )
) d1
,
(
select number,date from #date where date in
(select min(date) from #date group by convert(varchar(10),date,120) )
) d2
where convert(varchar(10),d1.date,120)=convert(varchar(10),d2.date,120)
number
-----------
2
3
