SQL 服务器 - 反向累计和

Question

我有一笔累计金额需要冲销。我如何在 SQL 服务器中执行此操作？

示例：

+---+-------------------+-----+
| id|               date|value|
+---+-------------------+-----+
| J1|2016-10-01 11:45:30|  100|
| J1|2016-10-02 11:30:30|  200|
| J1|2016-10-05 16:20:00|  400|
| J9|2016-10-06 08:35:00|  800|
| J9|2016-10-07 01:20:00|  900|
+---+-------------------+-----+

所需的数据帧：

+---+-------------------+-----+---------+
| id|               date|value|non_cum_value|
+---+-------------------+-----+---------+
| J1|2016-10-01 11:45:30|  100|        0|
| J1|2016-10-02 11:30:30|  200|      100|
| J1|2016-10-05 16:20:00|  400|      200|
| J9|2016-10-06 08:35:00|  800|      400|
| J9|2016-10-07 01:20:00|  900|      100|
+---+-------------------+-----+---------+

我的代码：

select t1.id, t1.value, DIFFERENCE(t1.value) as 'cum_sum'
from @t t1
inner join @t t2 on t1.id >= t2.id
group by t1.id, t1.value
order by t1.id

Answer 1

嗯嗯。 . .您似乎想从“上一个”行中减去该值。那将是：

select t.*,
       (t.value - lag(t.val) over (order by date)) as diff
from @t;

我不知道为什么你的标题中有“累计总和”。这并不能帮助我理解你想要做什么。

Answer 2

另一种可能性是：

SELECT *, V - SUM(V) OVER(ORDER BY D ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING)
FROM   T

假设测试数据为：

CREATE TABLE T (ID CHAR(2), D DATE, V FLOAT);

INSERT INTO T VALUES 
('J1', '2016-10-01 11:45:30',   100),
('J1', '2016-10-02 11:30:30',   200),
('J1', '2016-10-05 16:20:00',   400),
('J9', '2016-10-06 08:35:00',   800),
('J9', '2016-10-07 01:20:00',   900);

SQL 服务器 - 反向累计和

SQL Server - Reverse Cumulative Sum

sql

sql-server

cumulative-sum