--id为(1),(2)列行字段

select datadiff(dd,a.交易时间,b.交易时间)
from tableName a 
left join tableName b on a.id = b.id+1

------解决方案--------------------
SQL code
declare @t table(id int,d datetime,v float)
insert into @t select 1,'2008-01-05',4000
insert into @t select 2,'2008-01-03',2000
insert into @t select 3,'2008-01-02',1000

declare @利率 float
set @利率=0.03
select a.*,a.v*datediff(d,b.d,a.d)*@利率 from @t a left join @t b on a.id=b.id-1
/*
id          d                       v                      
----------- ----------------------- ---------------------- ----------------------
1           2008-01-05 00:00:00.000 4000                   240
2           2008-01-03 00:00:00.000 2000                   60
3           2008-01-02 00:00:00.000 1000                   NULL
*/

------解决方案--------------------
SQL code
create table tbf(id int,tim varchar(10),acc decimal(22,4))
insert into tbf values(1,'2009/01/01',40000)
insert into tbf values(2,'2009/02/01',30000)
insert into tbf values(3,'2009/03/01',20000)
insert into tbf values(4,'2009/04/01',10000)
insert into tbf values(5,'2009/04/07',60000)


select datediff(dd,a.tim,b.tim),a.*
from tbf a 
left join tbf b on a.id = b.id+1

drop table tbf
/*
            id          tim        acc
----------- ----------- ---------- ---------------------------------------
NULL        1           2009/01/01 40000.0000
-31         2           2009/02/01 30000.0000
-28         3           2009/03/01 20000.0000
-31         4           2009/04/01 10000.0000
-6          5           2009/04/07 60000.0000

(5 行受影响)

*/

------解决方案--------------------
SQL code
select datadiff(dd,a.交易时间,b.交易时间) from tableName a left join tableName b on a.id = b.id+1

------解决方案--------------------
探讨

a,b只是别名而已!


------解决方案--------------------
select datadiff(dd,a.交易时间,b.交易时间) from tableName a left join tableName b on a.id = b.id+1 


------解决方案--------------------
在插入下一条的时候,就把上一条记录的利率啥的就算了,增加个字段来记录.呵呵.不怕他利率变化,呵呵.

在每条上复杂些,但在查询上,大大简化,呵呵.

希望对你有帮助.

------解决方案--------------------
对的，这个只是别名  
楼上的解法就可以了  


SQL codecreate table tbf(id int,tim varchar(10),acc decimal(22,4))
insert into tbf values(1,'2009/01/01',40000)
insert into tbf values(2,'2009/02/01',30000)
insert into tbf values(3,'2009/03/01',20000)
insert into tbf values(4,'2009/04/01',10000)
insert into tbf values(5,'2009/04/07',60000)


select datediff(dd,a.tim,b.tim),a.*
from tbf a  
left join tbf b on a.id = b.id+1

drop table tbf
/*
           id          tim        acc
----------- ----------- ---------- ---------------------------------------
NULL        1           2009/01/01 40000.0000
-31         2           2009/02/01 30000.0000
-28         3           2009/03/01 20000.0000
-31         4           2009/04/01 10000.0000
-6          5           2009/04/07 60000.0000