每个日期期间合格客户的总购买百分比答案

【问题标题】：Percentage of total purchase for qualifying customers per date period每个日期期间合格客户的总购买百分比
【发布时间】：2020-07-09 08:58:45
【问题描述】：

我需要找出每个月购买巧克力的顾客的巧克力销量百分比。以下面的示例表为例，它只有一个 Month 条目：

Cust_Id     Customer    Quantity    Item        Amount in $     Date         Month and year
1           Maria       2           chocolate   10                 01/01/2020   1-2020
1           Maria       1           banana      5                  01/01/2020   1-2020
1           Maria       3           cookies     3                  01/01/2020   1-2020
2           Carlos      5           cookies     10                 01/01/2020   1-2020
2           Carlos      3           banana      3                  01/01/2020   1-2020
3           Jose        1           banana      5                  01/01/2020   1-2020
3           Jose        1           chocolate   3                  01/01/2020   1-2020
4           Anne        10          chocolate   20                 01/01/2020   1-2020
4           Anne        1           banana      5                  01/01/2020   1-2020
4           Anne        10          cookies     15                 01/01/2020   1-2020

在 2020 年 1 月购买了巧克力的客户（Maria、Jose 和 Anne，但不是 Carlos）中，巧克力的总购买量（百分比）有多少？

通过excel中的数据透视表解决：

Month   1 2020
Customer Name   SUM of Amount in $  Total amount of purchase    Overall %
Anne            20                  40  
Jose            3                   8   
Maria           10                  18  
Grand Total     33                  66                         50.00%

日期“1-2020”的解是50%

如何确定每个日期期间仅通过 SQL 购买巧克力的客户在每个日期期间售出的巧克力百分比？

mysql> select * from orders where date = '1-2020';
+----------+-------------+---------------+----------+-----------+--------+--------+
| order_id | customer_id | customer_name | quantity | item      | amount | date   |
+----------+-------------+---------------+----------+-----------+--------+--------+
|        1 |           1 | Maria         |        2 | chocolate |     10 | 1-2020 |
|        2 |           1 | Maria         |        1 | banana    |      5 | 1-2020 |
|        3 |           1 | Maria         |        3 | cookies   |      3 | 1-2020 |
|        4 |           2 | Carlos        |        5 | cookies   |     10 | 1-2020 |
|        5 |           2 | Carlos        |        3 | banana    |      3 | 1-2020 |
|        6 |           3 | Jose          |        1 | banana    |      5 | 1-2020 |
|        7 |           3 | Jose          |        1 | chocolate |      3 | 1-2020 |
|        8 |           4 | Anne          |       10 | chocolate |     20 | 1-2020 |
|        9 |           4 | Anne          |        1 | banana    |      5 | 1-2020 |
|       10 |           4 | Anne          |       10 | cookies   |     15 | 1-2020 |
+----------+-------------+---------------+----------+-----------+--------+--------+
10 rows in set (0.00 sec)

下面的查询几乎是正确的，只是它显示一月份的总额为 79 美元，而正确的金额是 66 美元（卡洛斯一月份没有买巧克力）

mysql> select date,     sum(case when item = 'chocolate' then amount end) as chocolate_amount,     sum(amount) as total_amount,     (sum(case when item = 'chocolate' then amount end) / sum(amount)) as percentage from orders where date = '1-2020' group by date having sum(case when item = 'chocolate' then 1 else 0 end) > 0;
+--------+------------------+--------------+------------+
| date   | chocolate_amount | total_amount | percentage |
+--------+------------------+--------------+------------+
| 1-2020 |               33 |           79 |     0.4177 |
+--------+------------------+--------------+------------+
1 row in set (0.00 sec)

同样，这里的正确计算应该是 33 / 66 = 50% 谢谢

【问题讨论】：

标签： sql netsuite domo

【解决方案1】：

您可以使用条件聚合：

select customer, 
       sum(case when item = 'chocolate' then amount end) as chocolate_amount,
       sum(amount) as total_amount,
       (sum(case when item = 'chocolate' then amount end) /
        sum(amount)
       ) as chocolate_ratio
from t
where date >= '2020-01-01' and date < '2020-02-01'
group by customer
having sum(case when item = 'chocolate' then 1 else 0 end) > 0

【讨论】：

谢谢，当表格有很多很多财政季度时，这会起作用吗？
另外，我需要找到总体百分比，即 1 月份的 50%
谢谢@Gordon Linoff，这个查询几乎给了我我想要的，除了这里的total_amount 包括每个时期没有购买巧克力的客户。我已经编辑了帖子并添加了一个示例。请问能再看看吗？谢谢
@mmenschig 。 . .我不明白你的问题。如果客户没有购买巧克力，则应通过having 子句过滤掉该客户。
我在下面添加了一个基于您提供的答案的答案。谢谢，很有帮助

【解决方案2】：

感谢 Gordon Linoff，我能够修改他的查询以提供所需的结果。虽然不是 100% 正确，但他的询问让我走上了正确的道路。

对于任何想知道查询和结果是什么的人：

查询

select o.date,
    sum(case when item = 'chocolate' then amount end) as chocolate_amount,
    t2.total_purchases,
    (sum(case when item = 'chocolate' then amount end) / t2.total_purchases) as percentage
from orders o
join (
    select date, sum(amount) as total_purchases
        from orders o1
        join (
            select distinct date, customer_id 
            from orders o2
            where item = 'chocolate'
        ) as chocolate_customers
using (date)
where o1.customer_id in (chocolate_customers.customer_id)
group by 1
) as t2 using (date)
group by date
having sum(case when item = 'chocolate' then 1 else 0 end) > 0;

结果

+--------+------------------+-----------------+------------+
| date   | chocolate_amount | total_purchases | percentage |
+--------+------------------+-----------------+------------+
| 1-2020 |               33 |              66 |     0.5000 |
| 2-2020 |               10 |              15 |     0.6667 |
| 3-2020 |               55 |              62 |     0.8871 |
+--------+------------------+-----------------+------------+
3 rows in set (0.00 sec)

再次感谢

【讨论】：