Question

我有如下示例数据。

Start_Dt    End_Dt     Dur_of_months  amount 
2020-01-01  2020-04-01   4             800

我有开始日期和结束日期以及几个月的持续时间。

将数量（800）除以4 = 200。

我希望每个月的总跑步次数为负数（200）。

输出：

mon_dt     amount 
Jan 2020    800
Feb 2020    600
Mar 2020    400
Apr 2020    200

我有一些代码可以增加开始日期和结束日期之间的月份

SELECT ID, DATE_FORMAT(Startdate + INTERVAL n.n MONTH, '%M %Y') AS Dates
FROM dates
JOIN (
  SELECT n10.n * 10 + n1.n * 1 AS n
  FROM (
    SELECT 0 n UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3
    UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6
    UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9
  ) n10
  CROSS JOIN (
    SELECT 0 n UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3
    UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6
    UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9
  ) n1
) n ON Startdate + INTERVAL n.n MONTH <= EndDate
ORDER BY ID, Startdate + INTERVAL n.n MONTH

如何在此添加运行总计。谁能建议我。

Answer 1

如果您正在运行MySQL 8.0，则可以对此使用递归查询：

with recursive cte as (
    select 
        start_dt, 
        end_dt, 
        dur_of_months, 
        1 lvl, 
        amount initial_amount,
        amount
    from mytable
    union all
    select
        start_dt + interval 1 month,
        end_dt,
        dur_of_months,
        lvl + 1,
        initial_amount,
        initial_amount * (1 - lvl / dur_of_months)
    from cte
    where start_dt < end_dt
)
select date_format(start_dt, '%M %Y') mon_dt, amount from cte order by start_dt

Demo on DB Fiddle ：

| mon_dt        | amount |
| ------------- | ------ |
| January 2020  | 800    |
| February 2020 | 600    |
| March 2020    | 400    |
| April 2020    | 200    |

在早期版本中，您将从现有查询开始：

select 
    date_format(start_dt + interval n.n month, '%M %Y') as mon_dt,
    amount * (1 - n / dur_of_months) amount
from mytable
join (
  select n10.n * 10 + n1.n * 1 as n
  from (
    select 0 n union all select 1 union all select 2 union all select 3
    union all select 4 union all select 5 union all select 6
    union all select 7 union all select 8 union all select 9
  ) n10
  cross join (
    select 0 n union all select 1 union all select 2 union all select 3
    union all select 4 union all select 5 union all select 6
    union all select 7 union all select 8 union all select 9
  ) n1
) n on start_dt + interval n.n month <= end_dt
order by start_dt + interval n.n month

Demo

如何在MYSQL中根据开始日期和结束日期获取运行总金额

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