超高分！MSSQLServer求相邻两行时间差的有关问题

超高分！MSSQLServer求相邻两行时间差的问题
表概况：
  .....交易时间...........当前余额.....
 .....(1)xxxx-xx-xx ........40000.0.....
 .....(2)xxxx-xx-xx.........30000.0.....
 .....(3)xxxx-xx-xx.........20000.0.....
 .....(4)xxxx-xx-xx.........10000.0.....

大概是这种情况，我想求相邻两行的 交易时间 的差，注意是相邻两行，然后用40000.0×时间(2)与时间(1)的差×日利率=利息。

请高手帮帮忙吧，小弟很急的，最好能有代码

[解决办法]

SQL code

--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
[解决办法]
SQL code
declare @t table(id int,d datetime,v float)insert into @t select 1,'2008-01-05',4000insert into @t select 2,'2008-01-03',2000insert into @t select 3,'2008-01-02',1000declare @利率 floatset @利率=0.03select 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                   2402           2008-01-03 00:00:00.000 2000                   603           2008-01-02 00:00:00.000 1000                   NULL*/
[解决办法]
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+1drop 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)*/
[解决办法]
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
[解决办法]
探讨

a,b只是别名而已!

[解决办法]
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 

[解决办法]
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.
[解决办法]
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 
 

(5 行受影响)

*/



[解决办法]
create table SalesItem
(
salesidint identity,
saledatedatetime,
balancemoney
)
insert into SalesItem(saledate,balance)
values('2009-1-1',899)
insert into SalesItem(saledate,balance)
values('2009-2-1',344)
insert into SalesItem(saledate,balance)
values('2009-2-21',333)
insert into SalesItem(saledate,balance)
values('2009-5-4',6665)
select * from SalesItem

select datediff(day,s1.saledate,S2.saledate) as balancedate
from SalesItem as S1
join SalesItem as S2
on  S2.salesid = S1.salesid + 1
[解决办法]
SQL code
with XX as(    select *, Row_Number() over(Order By [Date Column] desc) as index from [Table])select datediff(t1.[Date Column] - t2.[Date Column]) * t1.[Money] * [日利率] as [结果] from XX t1 inner join XX t2 on t1.index = t2.index + 1