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[英]Calculate Total Ending Quantity by using previous and next row value (LAG & LEAD) in SQL Server
[英]SQL Server : Using LAG() with calculated previous value
我在 SQL Server 2016 数据库表中有这样的数据:
时期 | PERIODE_FORECAST | 价值 |
---|---|---|
2021-08-01 | 2021-01-01 | 51384.673 |
2021-08-01 | 2021-02-01 | 44118.129 |
2021-08-01 | 2021-03-01 | 43164.446 |
2021-08-01 | 2021-04-01 | 38113.745 |
2021-08-01 | 2021-05-01 | 37306.956 |
2021-08-01 | 2021-06-01 | 38390.359 |
2021-08-01 | 2021-07-01 | 42692.390 |
2021-08-01 | 2021-08-01 | 39814.047 |
2021-08-01 | 2021-09-01 | 0.000 |
2021-08-01 | 2021-10-01 | 0.000 |
2021-08-01 | 2021-11-01 | 0.000 |
2021-08-01 | 2021-12-01 | 0.000 |
我有一个条件来填写零值,例如:
CASE
WHEN PERIODE_FORECAST > PERIODE
THEN [PREVIOUS_MONTH] * 0.1
ELSE VALUE
END
我正在尝试使用LAG()
函数,但它没有像我希望的那样出现。
询问 :
SELECT
PERIODE,PERIODE_FORECAST,
CASE
WHEN PERIODE_FORECAST > PERIODE
THEN LAG(VALUE, 1, 0) OVER (ORDER BY PERIODE_FORECAST ASC) * 0.1
ELSE VALUE
END VALUE
FROM
MyTable
结果 :
时期 | PERIODE_FORECAST | 价值 |
---|---|---|
2021-08-01 | 2021-01-01 | 51384.673 |
2021-08-01 | 2021-02-01 | 44118.129 |
2021-08-01 | 2021-03-01 | 43164.446 |
2021-08-01 | 2021-04-01 | 38113.745 |
2021-08-01 | 2021-05-01 | 37306.956 |
2021-08-01 | 2021-06-01 | 38390.359 |
2021-08-01 | 2021-07-01 | 42692.390 |
2021-08-01 | 2021-08-01 | 39814.047 |
2021-08-01 | 2021-09-01 | 3981.4047 |
2021-08-01 | 2021-10-01 | 0.000 |
2021-08-01 | 2021-11-01 | 0.000 |
2021-08-01 | 2021-12-01 | 0.000 |
预期成绩:
| PERIODE | PERIODE_FORECAST | VALUE |
| --------------|-----------------------|-----------|
| 2021-08-01 | 2021-01-01 | 51384.673 |
| 2021-08-01 | 2021-02-01 | 44118.129 |
| 2021-08-01 | 2021-03-01 | 43164.446 |
| 2021-08-01 | 2021-04-01 | 38113.745 |
| 2021-08-01 | 2021-05-01 | 37306.956 |
| 2021-08-01 | 2021-06-01 | 38390.359 |
| 2021-08-01 | 2021-07-01 | 42692.390 |
| 2021-08-01 | 2021-08-01 | 39814.047 |
| 2021-08-01 | 2021-09-01 | 3981.4047 |
| 2021-08-01 | 2021-10-01 | 398.14047 |
| 2021-08-01 | 2021-11-01 | 39.814047 |
| 2021-08-01 | 2021-12-01 | 3.9814047 |
|---------------|-----------------------|-----------|
有没有不创建存储过程的解决方案或解决方法?
谢谢
您的问题是,当您在第一个周期后 2 个月时,您的值滞后于 0。 我认为你必须在函数 lag() 的第二部分使用一个参数。 有点像: lag(VALUE, DATEDIFF(m,convert(date,'2021-08-01'),convert(date,'2021-10-01')) ,0 )
这里 :
滞后(值,DATEDIFF(m,转换(日期,PERIODE),转换(日期,PERIODE_FORECAST)),0)
您可能需要将此 datediff 添加为乘法器 0.1 的指数。
这是一个有趣的问题,我的方法可能不是最高效的,因为我是一名程序员,但它确实产生了您期望的结果。 我首先创建一个公共表表达式并使用 ROW_NUMBER 对行进行编号。 然后我回到 CTE 以找到具有最高 ROW_NUMBER 的单个 VALUE。 您可以使用 ROW_NUMBER 中的差异作为 10.0000 的 POWER 来获得乘数。
首先我创建了你的数据:
DECLARE @Forecast AS TABLE
(
[PERIODE] DATE NOT NULL,
[PERIODE_FORECAST] DATE NOT NULL,
[VALUE] DECIMAL(16, 8) NOT NULL
);
INSERT INTO @Forecast
(
[PERIODE],
[PERIODE_FORECAST],
[VALUE]
)
VALUES
('2021-08-01', '2021-01-01', 51384.673),
('2021-08-01', '2021-02-01', 44118.129),
('2021-08-01', '2021-03-01', 43164.446),
('2021-08-01', '2021-04-01', 38113.745),
('2021-08-01', '2021-05-01', 37306.956),
('2021-08-01', '2021-06-01', 38390.359),
('2021-08-01', '2021-07-01', 42692.390),
('2021-08-01', '2021-08-01', 39814.047),
('2021-08-01', '2021-09-01', 0.000),
('2021-08-01', '2021-10-01', 0.000),
('2021-08-01', '2021-11-01', 0.000),
('2021-08-01', '2021-12-01', 0.000);
一旦我有了这个,查询是:
;WITH [NumberedRows]
AS (SELECT [PERIODE],
[PERIODE_FORECAST],
[VALUE],
ROW_NUMBER() OVER (PARTITION BY [PERIODE]
ORDER BY [PERIODE_FORECAST]
) AS [rn]
FROM @Forecast)
SELECT [nr1].[PERIODE],
[nr1].[PERIODE_FORECAST],
CASE WHEN [sub].[VALUE] IS NOT NULL THEN
[sub].[VALUE] * POWER(10.0000000, ([sub].[rn] - [nr1].[rn]))
ELSE
[nr1].[VALUE]
END AS [VALUE]
FROM [NumberedRows] AS [nr1]
LEFT OUTER JOIN
(
SELECT TOP(1)
[nr2].[PERIODE], [nr2].[PERIODE_FORECAST], [nr2].[VALUE], [nr2].[rn]
FROM [NumberedRows] [nr2]
WHERE [nr2].[VALUE] <> 0
ORDER BY [nr2].[rn] DESC
) AS [sub]
ON [sub].[PERIODE] = [nr1].[PERIODE]
AND [sub].[rn] < [nr1].[rn];
结果:
时期 | PERIODE_FORECAST | 价值 |
---|---|---|
2021-08-01 | 2021-01-01 | 51384.673000 |
2021-08-01 | 2021-02-01 | 44118.129000 |
2021-08-01 | 2021-03-01 | 43164.446000 |
2021-08-01 | 2021-04-01 | 38113.745000 |
2021-08-01 | 2021-05-01 | 37306.956000 |
2021-08-01 | 2021-06-01 | 38390.359000 |
2021-08-01 | 2021-07-01 | 42692.390000 |
2021-08-01 | 2021-08-01 | 39814.047000 |
2021-08-01 | 2021-09-01 | 3981.404700 |
2021-08-01 | 2021-10-01 | 398.140470 |
2021-08-01 | 2021-11-01 | 39.814047 |
2021-08-01 | 2021-12-01 | 3.981405 |
这是间隙和孤岛问题,其中每个非零值都标志着新孤岛的开始。 分组后,您可以使用first_value
来挑选相应的“前一个值”,而row_number
用作计算乘数的偏移量。
with A as (
select *,
sum(case when "value" <> 0 then 1 else 0 end)
over (order by periode_forecast) as grp
from T
), B as (
select *,
first_value("value")
over (partition by grp order by periode_forecast) as pv,
power(10e, 1 - row_number()
over (partition by grp order by periode_forecast)) as mult
from A
)
select periode_forecast, "value", grp, pv, pv * mult as new_value
from B
order by periode_forecast;
您可能希望坚持使用decimal
数学并避免使用float
。 如果是这样,则调整power()
参考中的10e
。
如果没有有效的先前非零行,则结果将为零。 目前尚不清楚这是否会发生在您的数据中,或者如何以不同的方式对待它。
https://dbfiddle.uk/?rdbms=sqlserver_2016&fiddle=545d6e84ea43885788e45a2fb0393884
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