试试下面的逻辑-
此处演示

SELECT *,
SUM(A.money_step) over (
    partition by A.id_key, A.P 
    ORDER BY A.actual_date 
    ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
)as accumulated
FROM
(
    SELECT A.*,
    (
        SELECT CASE WHEN COUNT(*) >= 2 THEN 2 ELSE 1 END
        FROM your_table B 
        WHERE B.id_key = A.id_key
        AND B.actual_date <= A.actual_date
        AND attendance_status = 15
    ) P
    FROM your_table A
)A
ORDER BY A.id_key,A.actual_date

这是未经测试，因为我没有转录一个图像，然而，我认为这将工作。如果没有，则消耗品样品数据必须在此处显示：

WITH CTE AS
    (SELECT t.actual_date,
            t.id_key,
            t.attendance_status,
            t.money_step,
            ROW_NUMBER() OVER (PARTITION BY t.id_key,
                                            CASE attendance_status WHEN 15 THEN 1 ELSE 0 END
                               ORDER BY t.actual_date) AS RN
     FROM example t)
SELECT C.actual_date,
       C.id_key,
       C.attendance_status,
       C.money_step,
       SUM(t.money_step) OVER (PARTITION BY t.id_key, CASE WHEN RN > 1 THEN 1 ELSE 0 END
                               ORDER BY t.actual_date
                               ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS accumulated
FROM CTE C;

我强烈建议你只使用窗口函数。其实很简单：

SELECT t.*,
       SUM(t.money_step) OVER (PARTITION BY id_key, grp ORDER BY actual_date) as accumulated
FROM (SELECT t.*,
             SUM(CASE WHEN seqnum = 2 AND attendance_status = 15 THEN 1 ELSE 0 END) OVER
                 (PARTITION BY id_key ORDER BY actual_date) as grp
      FROM (SELECT t.*,
                   ROW_NUMBER() OVER (PARTITION BY id_key, attendance_status ORDER BY actual_date) as seqnum
            FROM t
           ) t
     ) t
ORDER BY t.id_key, t.actual_date;

这是一把小提琴。