用上限计算 SUM 并按比例乘以 SQL 中的系数答案

【问题标题】：Calculate SUM with cap and multiply pro-rata by coefficient in SQL用上限计算 SUM 并按比例乘以 SQL 中的系数
【发布时间】：2020-12-23 18:03:03
【问题描述】：

我的桌子或多或少看起来像这样：

+--------+---------------------+--------+------+
| CustId |      TransDate      | Amount | Coef |
+--------+---------------------+--------+------+
|      1 | 2020-09-04 11:02:00 |    400 |  0.1 |
|      1 | 2020-09-04 12:05:00 |    500 |  0.2 |
|      1 | 2020-09-04 13:02:00 |    400 |  0.1 |
|      1 | 2020-09-04 13:11:00 |    600 |  0.4 |
|      2 | 2020-09-04 10:01:00 |    300 |  0.3 |
|      2 | 2020-09-04 11:02:00 |    700 |  0.2 |
+--------+---------------------+--------+------+

我想要实现的是给每个客户一个“奖金”，计算为 Amount * Coef，但交易总额的上限为 1,500。 “红利”奖励给 SUM

例如，客户“1”将在前 3 笔交易中获得全额奖金，在第 4 笔交易中获得 1,500 的差额奖金。基本上，最终结果应该是这样的：

+--------+---------------------+--------+------+---------------------------------------------------+
| CustId |      TransDate      | Amount | Coef |                       Bonus                       |
+--------+---------------------+--------+------+---------------------------------------------------+
|      1 | 2020-09-04 11:02:00 |    400 |  0.1 | 40                                                |
|      1 | 2020-09-04 12:05:00 |    500 |  0.2 | 100                                               |
|      1 | 2020-09-04 13:02:00 |    400 |  0.1 | 40                                                |
|      1 | 2020-09-04 13:11:00 |    600 |  0.4 | 80 /*(this is given only for the remaining 200)*/ |
|      2 | 2020-09-04 10:01:00 |    300 |  0.3 | 90                                                |
|      2 | 2020-09-04 11:02:00 |    700 |  0.2 | 140                                               |
+--------+---------------------+--------+------+---------------------------------------------------+

提前致谢！

干杯。

【问题讨论】：

标签： sql sql-server ssms teradata-sql-assistant

【解决方案1】：

如果我理解正确，您需要一个累积和一些逻辑：

select t.*,
       (case when sum(amount) over (partition by custid order by transdate) <= 1500
             then coef * amount
             when sum(amount) over (partition by custid order by transdate) - amount < 1500
             then (1500 - sum(amount) over (partition by custid order by transdate)) as coef
             else 0
        end) as bonus
from t

【讨论】：

【解决方案2】：

[编辑]：现在使用 MAX 函数限制奖金金额。

数据

drop table if exists #tTEST;
go
select * INTO #tTEST from (values 
(1, '2020-09-04 11:02:00', 400, 0.1),
(1, '2020-09-04 12:05:00', 500, 0.2),
(1, '2020-09-04 13:02:00', 400, 0.1),
(1, '2020-09-04 13:11:00', 600, 0.4),
(1, '2020-09-05 13:11:00', 600, 0.4),
(2, '2020-09-04 10:01:00', 300, 0.3),
(3, '2020-09-04 11:02:00', 700, 0.2)) V(CustId, TransDate, Amount, Coef);

查询

declare @bonus_limit            int=1500;

with data_cte as (
    select *, sum(amount) over (partition by custid order by transdate) cum_amount,
            @bonus_limit-sum(amount) over (partition by custid order by transdate) cum_amount_diff
    from #tTEST t) 
select *, 
       (SELECT Max(v)  FROM (VALUES (0), (case when cum_amount_diff<0 then (Amount+cum_amount_diff)*Coef
                    else Amount*Coef end)) AS value(v)) Bonus
from data_cte;

结果

+--------+---------------------+--------+------+------------+-----------------+-------+
| CustId |      TransDate      | Amount | Coef | cum_amount | cum_amount_diff | Bonus |
+--------+---------------------+--------+------+------------+-----------------+-------+
|      1 | 2020-09-04 11:02:00 |    400 |  0.1 |        400 |            1100 |  40.0 |
|      1 | 2020-09-04 12:05:00 |    500 |  0.2 |        900 |             600 | 100.0 |
|      1 | 2020-09-04 13:02:00 |    400 |  0.1 |       1300 |             200 |  40.0 |
|      1 | 2020-09-04 13:11:00 |    600 |  0.4 |       1900 |            -400 |  80.0 |
|      1 | 2020-09-05 13:11:00 |    600 |  0.4 |       2500 |           -1000 |   0.0 |
|      2 | 2020-09-04 10:01:00 |    300 |  0.3 |        300 |            1200 |  90.0 |
|      3 | 2020-09-04 11:02:00 |    700 |  0.2 |        700 |             800 | 140.0 |
+--------+---------------------+--------+------+------------+-----------------+-------+

【讨论】：