求一简单经典语句。

[Haten]

TABLE A:
F_1 F_2
AA 1
BB 3
CC 7

如何得到以下结果：
F_1 F_2
AA 1
BB 2
CC 4

即：下一行减去上一行F_2的值。

[wzy_love_sly]

declare @tb table (f1 varchar(10),f2 int)
insert into @tb select 'aa',1
insert into @tb select 'bb',3
insert into @tb select 'cc',7
select a.f1,a.f2-isnull(b.f2,0) as '差' from (
select *, px=row_number() over(order by f1)from @tb )a
left join(
select *, px=row_number() over(order by f1)from @tb )b
on a.px=b.px+1

f1 差
aa 1
bb 2
cc 4

[pt1314917]

declare @a table(F_1 varchar(10),F_2 int) 
insert into @a select 'AA',       1 
insert into @a select 'BB',       3 
insert into @a select 'CC',       7 


select id=identity(int,1,1),* into #temp from @a


select a.f_1,isnull(a.f_2-b.f_2,a.f_2)f_2 from #temp a left join #temp b
on a.id=b.id+1 

[dawugui]

create table tb(F_1 varchar(10) , F_2 int)
insert into tb values('AA', 1 )
insert into tb values('BB', 3 )
insert into tb values('CC', 7 )
go

select tb.* from tb where f_1 in (select min(f_1) from tb)
union all
select t2.f_1 , f_2 = t2.f_2 - t1.f_2 from
(SELECT * , px = (SELECT COUNT(f_1) FROM tb WHERE f_1 < t.f_1) + 1 FROM tb t) t1,
(SELECT * , px = (SELECT COUNT(f_1) FROM tb WHERE f_1 < t.f_1) + 1 FROM tb t) t2
where t1.px = t2.px - 1

drop table tb

/*
F_1        F_2         
---------- ----------- 
AA         1
BB         2
CC         4

（所影响的行数为 3 行）
*/

[chengg0769]

拜托楼主。sql的表里面没有行这个概念。

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