求助！！高手请进~~~~100分奉送！！！

[heiguangbao]

我现在遇到一个比较严峻的问题，请高手解答
问题：字段a 字段b
1 7
3 8
2 4
5 9
实现的目的是：（字段b的第一条记录-字段a的第一条记录）/字段b的第一条记录，
（字段b的前两条记录-字段a的前两条记录）/字段b的前两条记录
一直这样循环下去
例如：(7-1）/7,((7+8)-(1+3))/(7+8),((7+8+4)-(1+3+2))/(7+8+4),
((7+8+4+9)-(1+3+2+5))/(7+8+4+9)
能用自定义函数实现吗？或者其他方法，先谢谢大家了！

[CrazyFor]

declare @i numeric(28,6),@j numeric(28,6),@V numeric(28,6)
set @i=0,@j=0
select *,0 as ValueCol into #temp from 表
update #temp
set ValueCol=((字段b+@j)-(字段a+@i))/(字段b+@j)
,@I=@i字段a
,@j=@j字段b
select * from #temp

[CrazyFor]

更正一下，
declare @i numeric(28,6),@j numeric(28,6)
set @i=0
set @j=0
select *,case(0 as numeric(28,6)) as ValueCol into #temp from 表
update #temp
set ValueCol=((字段b+@j)-(字段a+@i))/(字段b+@j)
,@I=@i字段a
,@j=@j字段b
select * from #temp

[pengdali]

select identity(int,1,1) id,* from 你的表

select ((select sum(b) from 你的表 where id<=tem.id)-(select sum(a) from 你的表 where id<=tem.id))/(select sum(b) from 你的表 where id<=tem.id) 结果 from 表

[pengdali]

select identity(int,1,1) id,* from 你的表

select ((select sum(b) from 你的表 where id<=tem.id)-(select sum(a) from 你的表 where id<=tem.id))/(select sum(b) from 你的表 where id<=tem.id) 结果 from 你的表 tem

[Methodor]

DBMS:通过临时表或游标可实现~
PB:利用datawindow实现起来更简单方便~

[happydreamer]

select *,cast(0 as decimal(10,2)) as v into #r from a


declare @i decimal(10,4), @j decimal(10,4)
select @i=0,@j=0

update #r
set @i=@i+a,@j=@j+b,v=(@j-@i)/ @j

[hjb111]

create proc addab1
as
declare @aa int,@bb int
declare @sumab int,@addb int
declare @diva real
set @sumab=0
set @addb=0
declare a cursor for select a,b from addab

open a
fetch next from a into @aa,@bb
while @@fetch_status=0
begin
set @sumab=@sumab+@aa
set @addb=@addb+@bb
set @diva=(@addb-@sumab)/@addb
fetch next from a into @aa,@bb
end

close a
deallocate a

[愉快的登山者]

select *, identity(int,1,1) as id into #t from yourtable

select a, b, (select (sum(a) - sum(b)) / sum(b) from #t where id <= A.id) result from #t as A

[愉快的登山者]

select *, identity(int,1,1) as id into #t from yourtable

select a, b, (select (sum(cast(a as numeric(8,4)) - sum(cast(b as numeric(8,4))) / sum(cast(b as numeric(8,4)) from #t where id <= A.id) result from #t as A

[hjb111]

select *, identity(int,1,1) as id into #t from addab

select a, b, (select (sum(cast(b as float)) - sum(cast(a as float))) / sum(cast(b as float)) from #t

where id <= A.id) result from #t as A