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", "2020-01-03", "500", "0"),
("NL", "2020-01-03", "320", "0"),
("FR", "2020-01-03", "350", "0"),
("None", "2020-01-30", "0", "2000"),

("DE", "2020-02-15", "700", "0"),
("NL", "2020-02-15", "420", "0"),
("FR", "2020-02-15", "180", "0"),
("None", "2020-02-29", "0", "5000"),

("DE", "2020-03-27", "180", "0"),
("NL", "2020-03-27", "670", "0"),
("FR", "2020-03-27", "970", "0"),
("None", "2020-03-31", "0", "4000");

预期结果：

sales_date      country       sales_volume     fix_costs
2020-01-03        DE              500           37.95  (=2000/31 = 64.5 * 0.59)
2020-01-03        FR              350           26.57  (=2000/31 = 64.5 * 0.41)
2020-01-03        NL              320            0.00
2020-02-15        DE              700          137.15  (=5000/29 = 172.4 * 0.8)   
2020-02-15        FR              180           35.27  (=5000/29 = 172.4 * 0.2)  
2020-02-15        NL              420            0.00    
2020-03-27        DE              180           20.20  (=4000/31 = 129.0 * 0.16) 
2020-03-27        FR              970          108.81  (=4000/31 = 129.0 * 0.84)   
2020-03-27        NL              670            0.00

关于this question中的解决方案，我目前使用以下查询将monthly fix_costs每天划分到每个国家：

select 
    sales_date, 
    country, 
    sum(sales_volume),
    sum(sum(fix_costs)) over(partition by year(sales_date), month(sales_date))
        / day(last_day(sales_date)) 
        * sum(sales_volume)
        / sum(sum(sales_volume)) over(partition by sales_date)
        as fix_cost_per_day
from sales
group by 1,2;

一切正常。

但是，现在我希望该国家NL被排除在share_calculation之外，并且始终保持0，如您在预期结果中所见。我该如何修改查询才能实现此目的？

Answer 1

您可以使用case表达式，既可以在计算中使用，也可以在窗口sum内使用：

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)
    else 0
    end as fix_cost_per_day
from sales
group by 1,2;

Demo on DB Fiddle ：

sales_date | country | sum(sales_volume) | fix_cost_per_day
:--------- | :------ | ----------------: | ---------------:
2020-01-03 | DE      |               500 |      37.95066414
2020-01-03 | FR      |               350 |      26.56546490
2020-01-03 | NL      |               320 |             null
2020-01-30 | None    |                 0 |             null
2020-02-15 | DE      |               700 |     137.14733542
2020-02-15 | FR      |               180 |      35.26645768
2020-02-15 | NL      |               420 |             null
2020-02-29 | None    |                 0 |             null
2020-03-27 | DE      |               180 |      20.19635344
2020-03-27 | FR      |               970 |     108.83590463
2020-03-27 | NL      |               670 |             null
2020-03-31 | None    |                 0 |             null

Answer 2

只需使用条件聚合：

select 
    sales_date, 
    country, 
    sum(sales_volume),
    (case when country = 'NL' then 0
          else sum(sum(case when country = 'NL' then 0 else fix_costs end)) over(partition by year(sales_date), month(sales_date)) / day(last_day(sales_date)) *
               sum(sales_volume) /
               sum(sum(sales_volume)) over(partition by sales_date)
     end)  as fix_cost_per_day
from sales
group by 1,2;

根据每日收入分成将固定值分配给各个国家/地区时，排除某些值

2 个答案: