根据收入份额将每月价值拆分为天数和国家/地区

Question

DB-Fiddle:

CREATE TABLE sales (
    id int auto_increment primary key,
    country VARCHAR(255),
    sales_date DATE,
    sales_volume INT,
    fix_costs INT
);

INSERT INTO sales
(country, sales_date, sales_volume, fix_costs
)
VALUES 

("DE", "2020-12-01", "500", "0"),
("NL", "2020-12-01", "320", "0"),
("FR", "2020-12-01", "350", "0"),

("DE", "2020-12-02", "700", "0"),
("NL", "2020-12-02", "420", "0"),
("FR", "2020-12-02", "180", "0"),

("DE", "2020-12-03", "0", "0"),
("NL", "2020-12-03", "0", "0"),
("FR", "2020-12-03", "0", "0"),

("None", "2020-12-31", "0", "2000");

预期结果：

sales_date   |   country    |   sales_volume   |     fix_costs
-------------|--------------|------------------|------------------------------------------
2020-12-01   |     DE       |       500        |     37.95  (=2000/31 = 64.5 * 0.59)
2020-12-01   |     FR       |       350        |     26.57  (=2000/31 = 64.5 * 0.41)
2020-12-01   |     NL       |       320        |      0.00
-------------|--------------|------------------|------------------------------------------
2020-12-02   |     DE       |       700        |     51.32  (=2000/31 = 64.5 * 0.8)   
2020-12-02   |     FR       |       180        |     13.20  (=2000/31 = 64.5 * 0.2)  
2020-12-02   |     NL       |       420        |      0.00
-------------|--------------|------------------|------------------------------------------    
2020-12-03   |     DE       |         0        |     32.26  (=2000/31 = 64.5 * 0.5) 
2020-12-03   |     FR       |         0        |     32.26  (=2000/31 = 64.5 * 0.5)    
2020-12-03   |     NL       |         0        |      0.00
-------------|--------------|------------------|-------------------------------------------

---

在上面的结果中，我想根据 daily 将每月的 fix_cost 值 (in this case 2000) 拆分为国家 DE 和 FR 每天 > sales_volume 在所选countries 的每个 中。到目前为止，我已经使用了这个question 的解决方案，效果很好：

SELECT
    sales_date, 
    country, 
    SUM(sales_volume),
    (CASE WHEN country <> 'NL'
        THEN SUM(SUM(fix_costs)) over(partition BY year(sales_date), month(sales_date))
            / day(last_day(sales_date)) 
            * SUM(sales_volume)
            / SUM(CASE WHEN country <> 'NL' THEN SUM(sales_volume) ELSE 0 END) over(partition BY sales_date)
    END) AS fix_cost_per_day
FROM sales
GROUP BY 1,2;

---

但是，正如您在 2020-12-03 上的数据中看到的那样，根本没有销售。
因此，上面的查询无法在这一天将每日fix_costs 拆分到每个国家/地区。
为了解决这个问题，我想修改查询，这样万一有一天没有sales_volume，值应该被拆分50/50。

我必须如何修改查询才能实现这一点？

Answer 1

如果我理解正确，你。想在每个月的每一天分配 2,000 个，并分配给除'NL' 以外的所有国家/地区。

这是带有窗口函数的算术：

• 使用last_day() 获取当月的天数。
• 使用窗口函数计算除'NL' 之外的每个国家/地区的收入比例。
• 使用条件逻辑，所以'NL' 是特殊的。

作为SELECT 查询：

select s.*,
       (case when country = 'NL' then 0
             else ((2000 / day(last_day(sales_date))) *
                   sales_volume / nullif(sum(case when country <> 'NL' then sales_volume end) over (partition by sales_date), 0)
                  )
        end) as imputed_fixed_costs
from sales s
order by id;

注意：2000 使代码更清晰，但显然您可以使用以下命令将其输入到查询中：

from sales s cross join
     (select sum(fix_costs) as fixed_costs
      from sales
      where country = 'None'
     ) f

Here 是一个 dbfiddle。

如果您真的想更改数据，这很容易合并到update，尽管问题中的示例查询是select。

处理没有销售的月份：

select s.*,
       (case when country = 'NL'
             then 0
             when sum(case when country <> 'NL' then sales_volume end) over (partition by sales_date) > 0
             then ((2000 / day(last_day(sales_date))) *
                   sales_volume / nullif(sum(case when country <> 'NL' then sales_volume end) over (partition by sales_date), 0)
                  )
             else (2000 / day(last_day(sales_date))) * 1 / sum(country <> 'NL') over (partition by sales_date)
        end) as imputed_fixed_costs
from sales s
order by id;