列'金额'第一行的值为5,第二行的值为-10。
有没有办法让oracle的sum(amount)over()函数为第二行返回0而不是-5?
答案 0 :(得分:2)
公然使用Rajesh Chamarthi的示例来源:但改变以显示更多的消极和积极的......并显示案例如何将所有负面变为零,同时保持其他金额......
with t as (
select 5 as x, 1 as id from dual
union all
select -10, 2 as id from dual
union all
select 7, 3 as id from dual
union all
select -5, 4 as id from dual
union all
select -2, 5 as id from dual
),
B as (select t.x,
case when sum(x) over (order by id) < 0 then 0
else sum(x) over (order by id)
end Amount
from t
order by id)
Select X, Case when amount < 0 then 0 else amount end as Amount from B;
T Amount
5 5
-10 0
7 2
-5 0
-2 0
----尝试2(第一次尝试保留为下面的评论引用它)
当数量低于0时,我无法弄清楚如何中断窗口函数将值重置为0 ...所以我使用了一个递归CTE,这让我有了更大的控制权。
如果id不是顺序的,我们可以添加一个row_Number,这样我们就可以加入一个ID ......或者我们可以使用min(),其中&gt; OLDID。我假设我们有一个唯一的密钥唯一ID或某种方式按照您希望总和发生的顺序“排序”记录......
with aRaw as (
select 5 as x, 15 as id from dual
union all
select -10, 20 as id from dual
union all
select 7, 32 as id from dual
union all
select 2, 46 as id from dual
union all
select -15, 55 as id from dual
union all
select 3, 68 as id from dual
),
t as (Select A.*, Row_number() over (order by ID) rn from aRAW A),
CTE(RN, ID, x, SumX) AS (
Select T.RN, T.ID, x, X from t WHERE ID = (Select min(ID) from t)
UNION ALL
Select T.RN, T.ID, T.X, case when T.X+CTE.SumX < 0 then 0 else T.X+Cte.sumX end from T
INNER JOIN CTE
on CTE.RN+1=T.RN)
Select * from cte;
答案 1 :(得分:1)
您可以使用案例陈述,但这不是真正的总计
with t as (
select 5 as x, 1 as id from dual
union all
select -10, 2 as id from dual
union all
select 20, 3 as id from dual
union all
select 30, 4 as id from dual
union all
select 10, 5 as id from dual
)
select t.x,
case when sum(x) over (order by id) < 0 then 0
else sum(x) over (order by id)
end running_total
from t
order by id;
X RUNNING_TOTAL
5 5
-10 0
20 15
30 45
10 55