急急急呀，关于sql行交叉的计算？？？？

Question

r3 計算公式(r1-r2)-r2 ？

Answer 1

declare @T table([id] int,[r1] int,[r2] int)
Insert @T
select 1,5,0 union all
select 2,0,2 union all
select 3,0,1 union all
select 4,0,1 union all
select 5,0,1
 
Select 
*,(select sum([r1]-[r2]) from @T where [id]<=t.[id])r3
from 
@T t(5 個資料列受到影響)
id          r1          r2          r3
----------- ----------- ----------- -----------
1           5           0           5
2           0           2           3
3           0           1           2
4           0           1           1
5           0           1           0(5 個資料列受到影響)

Answer 2

declare @t1 table(id int, sr numeric(12,2),zc numeric(12,2))
insert into @t1 
select 1,5,0 union all
select 2,0,2 union all
select 3,0,1 union all
select 4,0,1 union all
select 5,0,1 SELECT id,sr,zc,(select sum(a.sr-a.zc) from @t1 a where a.id<=b.id) FROM @t1 b(5 行受影响)
id          sr                                      zc                                      
----------- --------------------------------------- --------------------------------------- ---------------------------------------
1           5.00                                    0.00                                    5.00
2           0.00                                    2.00                                    3.00
3           0.00                                    1.00                                    2.00
4           0.00                                    1.00                                    1.00
5           0.00                                    1.00                                    0.00(5 行受影响)

Answer 3

(r1-r2)-r2，可能不对，但是要实现这种效果，r3要计算出来

Answer 4

declare @t table (id int, r1 int , r2 int) 
insert @t  select 1,  5,  0 union all select  
2,  0,  2  union all select 
3,  0,  1  union all select 
4,  0,  1  union all select 
5,  0,  1 
Select 
    *,(select sum([r1]-[r2]) from @T where [id]<=t.[id])r3
from 
    @T t