[英]SQL Server - Reverse Cumulative Sum
我有一個需要沖銷的累計金額。 如何在 SQL Server 中執行此操作?
例子:
+---+-------------------+-----+
| 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|
+---+-------------------+-----+
所需的數據幀:
+---+-------------------+-----+---------+
| 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|
+---+-------------------+-----+---------+
我的代碼:
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
嗯嗯。 . . 您似乎想從“上一個”行中減去該值。 那將是:
select t.*,
(t.value - lag(t.val) over (order by date)) as diff
from @t;
我不確定你為什么在標題中有“累計金額”。 這無助於我理解你想要做什么。
另一種可能性是:
SELECT *, V - SUM(V) OVER(ORDER BY D ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING)
FROM T
假設測試數據為:
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);
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