两条数据间的时间

start_dt id project
2011-12-02 23:30:00 1 NM1
2011-12-05 22:30:00 2 NM2
2011-12-02 14:45:00 3 NM1

如何根据以上的条件，查询每个project的用时。
如project NM2用时等于 2011-12-05 22:30:00 减去 2011-12-02 23:30:00
project NM1用时等于 2011-12-02 23:30:00 减去 2011-12-02 07:00:00（查询开始时间）。

结果是

use_dt project
？？ NM2
？？ NM1

请大侠们帮帮忙。。。
感激不尽！！

select t.* , dateadd(ss,isnull((select top 1 start_dt from tb where id < t.id order by id) , '2011-12-02 07:00:00'),t.start_dt) from tb t

SQL code

create table tb(start_dt datetime,id int,project varchar(10))
insert into tb values('2011-12-02 23:30:00', 1 ,'NM1')
insert into tb values('2011-12-05 22:30:00', 2 ,'NM2')
insert into tb values('2011-12-02 14:45:00', 3 ,'NM1')
go

select t.* , [用时（s）] = abs(datediff(ss,isnull((select top 1 start_dt from tb where id < t.id order by id) , '2011-12-02 07:00:00'),t.start_dt)) from tb t

drop table tb

/*
start_dt                                               id          project    用时（s）       
------------------------------------------------------ ----------- ---------- ----------- 
2011-12-02 23:30:00.000                                1           NM1        59400
2011-12-05 22:30:00.000                                2           NM2        255600
2011-12-02 14:45:00.000                                3           NM1        31500

（所影响的行数为 3 行）

*/

SQL code

create table tb(start_dt datetime,id int,project varchar(10))
insert into tb values('2011-12-02 23:30:00', 1 ,'NM1')
insert into tb values('2011-12-05 22:30:00', 2 ,'NM2')
insert into tb values('2011-12-02 14:45:00', 3 ,'NM1')
go

select t.* , [用时（s）] = abs(datediff(ss,isnull((select top 1 start_dt from tb where id < t.id order by id) , '2011-12-02 07:00:00'),t.start_dt)) from tb t
/*
start_dt                                               id          project    用时（s）       
------------------------------------------------------ ----------- ---------- ----------- 
2011-12-02 23:30:00.000                                1           NM1        59400
2011-12-05 22:30:00.000                                2           NM2        255600
2011-12-02 14:45:00.000                                3           NM1        31500

（所影响的行数为 3 行）

*/

select project , sum([用时（s）]) [用时（s）] from
(
  select t.* , [用时（s）] = abs(datediff(ss,isnull((select top 1 start_dt from tb where id < t.id order by id) , '2011-12-02 07:00:00'),t.start_dt)) from tb t
) m
group by project
/*
project    用时（s）       
---------- ----------- 
NM1        90900
NM2        255600

（所影响的行数为 2 行）
*/

drop table tb

SQL code

if object_id('[TB]') is not null drop table [TB]
go
create table [TB] (start_dt datetime,id int,project nvarchar(6))
insert into [TB]
select '2011-12-02 23:30:00',1,'NM1' union all
select '2011-12-05 22:30:00',2,'NM2' union all
select '2011-12-02 14:45:00',3,'NM1'

select * from [TB]


SELECT B.project,ABS(DATEDIFF(ss,B.start_dt,A.start_dt)) AS 用时
FROM TB A
INNER JOIN TB B ON A.id = B.id - 1

/*
project    用时
NM2    255600
NM1    287100*/

SQL code

select 
  b.project,abs(datediff(ss,b.start_dt,a.start_dt))
from
  a,b
where
   a.id=b.id-1