根据前面的公式计算电流

Question

我有一个场景，当前列值是根据公式计算的先前值计算的 组首行没有之前的值所以不考虑

损失公式= relase-withdraw-least(previous_row_loss,reverse)

下面损失是我需要计算的列。

我尝试了以下查询，但没有得到预期的输出。你能在这里指导我吗？

SELECT
pid,release,withdraw,reverse,
SUM(release - withdraw - LEAST( LAG(loss,1,0) OVER (ORDER BY pid)),reverse))  as loss 
FROM transactions

Answer 1

您可以使用 MODEL 子句：

SELECT *  
FROM   (
  SELECT t.*,
         ROW_NUMBER() OVER (ORDER BY pid, fiscalperiod) AS rn
  FROM   table_name t
)
MODEL
  DIMENSION BY (rn)
  MEASURES (pid, fiscalperiod, release, withdraw, reverse, 0 AS loss)
  RULES (
    loss[1]    = release[1] - withdraw[1] - reverse[1],
    loss[rn>1] = release[cv()] - withdraw[cv()] - LEAST(reverse[cv()], loss[cv()-1])
                 + loss[cv()-1]
);

或者，递归查询的效率可能低得多：

WITH numbered_rows AS (
  SELECT t.*,
         ROW_NUMBER() OVER (ORDER BY pid, fiscalperiod) AS rn
  FROM   table_name t
),
recursive_query (rn, pid, fiscalperiod, release, withdraw, reverse, loss) AS (
  SELECT rn,
         pid,
         fiscalperiod,
         release,
         withdraw,
         reverse,
         release - withdraw - reverse
  FROM   numbered_rows
  WHERE  rn = 1
UNION ALL
  SELECT n.rn,
         n.pid,
         n.fiscalperiod,
         n.release,
         n.withdraw,
         n.reverse,
         n.release - n.withdraw + GREATEST(r.loss - n.reverse, 0)
  FROM   numbered_rows n
         INNER JOIN recursive_query r
         ON (n.rn = r.rn + 1)
)
SELECT *
FROM   recursive_query;

其中，对于您的示例数据：

CREATE TABLE table_name (pid, fiscalperiod, release, withdraw, reverse) AS
SELECT 'A1', 2022001, 10, 10, 10 FROM DUAL UNION ALL
SELECT 'A1', 2022002, 20, 13,  2 FROM DUAL UNION ALL
SELECT 'A1', 2022003, 20, 20, 10 FROM DUAL UNION ALL
SELECT 'A2', 2022002, 15, 10, 13 FROM DUAL;

双输出：

RN	PID	FISCALPERIOD	RELEASE	WITHDRAW	REVERSE	LOSS
1	A1	2022001	10	10	10	-10
2	A1	2022002	20	13	2	7
3	A1	2022003	20	20	10	0
4	A2	2022002	15	10	13	5

db<>fiddle here