[英]Calculate column based on previous value of the same calculated column in MySQL
[英]Calculate column value based on previous row and the same calculated column
我正在努力根据Val1
和Val2
计算Calc
。
Calc
= previous_row.Calc
+ previousr_row.Val1
- previous_row.Val2
输入数据按Date
排序。
预期 output:
+---------+--------+------------+------+
| Val1 | Val2 | Date | Calc |
+---------+--------+------------+------+
| 0,00 | 0,00 | 2016-01-01 | 0 |
| 1000,00 | 0,00 | 2020-01-01 | 0 |
| 0,00 | 0,00 | 2020-01-15 | 1000 |
| 0,00 | 500,00 | 2020-02-01 | 1000 |
| 0,00 | 300,00 | 2020-03-01 | 500 |
| 0,00 | 0,00 | 2020-03-15 | 200 |
| 0,00 | 200,00 | 2020-04-01 | 200 |
+---------+--------+------------+------+
已经尝试过 LAG function,成功从上一行获取数据,但我无法从上一行获取 Calc 计算值:
LAG(Val1) OVER (ORDER By Date) - LAG(Val2) OVER (ORDER BY Date)
在现实世界的场景中,我将添加 PARTITION BY,但这是不同的故事。 现在想保持简单。
更新:受他人启发:
SUM(Val1) OVER(ORDER BY Data) - SUM(Val2) OVER(ORDER BY Data) - Val1 + Val2 AS Calc
虽然它计算了正确的值,但这是否有效?
我正在使用最新的 SQL 服务器 2019 / Azure SQL。
我认为您正在寻找累积和函数:
select t.*,
max(val1) over (order by date) - sum(val2) over (order by date)
from t;
ALTER FUNCTION calc (@lagDate DATE)
RETURNS INT
AS
BEGIN
IF @lagDate IS NULL
RETURN 0
DECLARE @r INT
SELECT @r = isnull(lag(val1) OVER ( ORDER BY [date] ), 0) -
isnull(lag(val2) OVER ( ORDER BY [date] ), 0) +
dbo.calc(lag([date]) OVER ( ORDER BY [date] ))
FROM dbo.ss
WHERE [date] <= @lagDate
RETURN @r
END
GO
SELECT *
,dbo.calc([date]) calc
FROM dbo.ss
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