两条数据间的时间解决方案

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！！

[解决办法]
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
[解决办法]

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')goselect 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 tdrop table tb/*start_dt                                               id          project    用时（s）       ------------------------------------------------------ ----------- ---------- ----------- 2011-12-02 23:30:00.000                                1           NM1        594002011-12-05 22:30:00.000                                2           NM2        2556002011-12-02 14:45:00.000                                3           NM1        31500（所影响的行数为 3 行）*/
[解决办法]
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')goselect 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        594002011-12-05 22:30:00.000                                2           NM2        2556002011-12-02 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) mgroup by project/*project    用时（s）       ---------- ----------- NM1        90900NM2        255600（所影响的行数为 2 行）*/drop table tb
[解决办法]
SQL code
if object_id('[TB]') is not null drop table [TB]gocreate table [TB] (start_dt datetime,id int,project nvarchar(6))insert into [TB]select '2011-12-02 23:30:00',1,'NM1' union allselect '2011-12-05 22:30:00',2,'NM2' union allselect '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 AINNER JOIN TB B ON A.id = B.id - 1/*project    用时NM2    255600NM1    287100*/