如何结合 SELECT 语句来计算 SQL Server 中的百分比、成功和失败？

Question

想象一张桌子：

CUST_PROMO (customer_id,PROMOTION) 用作客户收到的每个促销之间的映射。

select promotion, count(customer_id) as promo_size
from CUST_PROMO
group by promotion

这让我们得到每次促销活动的客户总数。

现在，我们有了 CUSTOMER (customer_id, PROMO_RESPONDED,PROMO_PURCHASED)，它列出了客户以及哪些促销活动让客户做出了回应，以及哪些促销活动让他们购买了。

select PROMO_RESPONDED, count(customer_id) as promo_responded 
from CUSTOMER
group by PROMO_RESPONDED

select PROMO_PURCHASED,count(customer_id) as promo_responded 
from CUSTOMER 
group by PROMO_PURCHASED

这一切都是不言自明的；现在我得到了每个促销活动成功的人数。

但是;我想结束的是[以 CSV 形式]

PROMOTION,PROMO_SIZE,PROMO_RESPONDED,PROMO_PURCHASED,PROMO_RESPSUCCESSRATE,blah

1,100,12,5,12%,...
2,200,23,14,11.5%,...

我不知道该怎么做。我可以联合上面的三个查询；但这实际上并没有得到我想要的结果。我考虑过创建一个内存表，插入每个促销值，然后执行一个更新语句并对其进行连接以设置每个值——但这很麻烦；并且需要为每个表/选择语句创建一个新的 UPDATE 语句。我还可以为每个结果集制作一个临时表，然后将它们连接在一起；但真的；谁愿意这样做？

我想不出任何有意义的方法来加入这些数据；因为我正在处理聚合。

所以，充其量，我需要一个函数，就像 UNION 一样，将组合结果集，但实际上会组合键上的类似列并添加这些列，而不是添加行的联合。描述使它听起来像一个 JOIN；但我看不到那个工作。

感谢您的帮助！

Answer 1

这行得通吗？我不确定除法和乘法运算符，但我相信我的逻辑很好。关键是在 select 语句中使用相关的 select 子语句。

SELECT c.promotion, 
       COUNT(c.customer_id) as promo_size, 
       (SELECT COUNT(customer_id)
          FROM CUSTOMER
         WHERE PROMO_RESPONDED = c.promotion) PROMO_RESPONDED,
       (SELECT COUNT(customer_id)
          FROM CUSTOMER
         WHERE PROMO_PURCHASED = c.promotion) PROMO_PURCHASED,
       (SELECT COUNT(customer_id) *100/count(c.customer_id)
         FROM CUSTOMER
        WHERE PROMO_RESPONDED = c.promotion)                                                      
FROM CUST_PROMO c
GROUP BY c.promotion

使用解码的更简洁的解决方案。仍然不确定数学是否有效

select PROMOTION, count(CUSTOMER_ID) as promo_size, 
       SUM(DECODE(PROMO_RESPONDED, PROMOTION, 1, 0)) PROMO_RESPONDED,
       SUM(DECODE(PROMO_PURCHASED, PROMOTION, 1, 0)) PROMO PURCHASED,
       SUM(DECODE(PROMO_RESPONDED, PROMOTION, 1, 0))*100/count(CUSTOMER_ID) PROMO_RESPONDED
from CUST_PROMO join CUSTOMER using CUSTOMER_ID
group by PROMOTION 

Answer 2

WITH tmp AS
(
    SELECT  PROMOTION, 0 as promo_responded, 0 as promo_purchased, COUNT(customer_id) as total
    FROM    CUST_PROMO
    GROUP BY PROMOTION
    SELECT  PROMOTION, COUNT(customer_id) as promo_responded, 0 as promo_purchased, 0 as total
    FROM    CUSTOMER
    GROUP BY PROMO_RESPONDED
    UNION   
    SELECT  PROMOTION, COUNT(customer_id) as promo_purchased, 0 as promo_responded, 0 as total
    FROM    CUSTOMER
    GROUP BY PROMO_PURCHASED
)
SELECT  PROMOTION, SUM(promo_responded) as TotalResponded, SUM(promo_purchased) as TotalPurchased, SUM(Total) as TotalSize,
        SUM(promo_responded)/SUM(Total) as ResponseRate, SUM(promo_purchased)/SUM(Total) as PurchaseRate
FROM    tmp

Answer 3

是的，我认为JOINing 三个聚合查询是要走的路。 LEFT JOINs 的存在是为了以防某些促销活动没有回应或没有购买。

我还将COUNT(customer_id) 更改为COUNT(*)。结果是一样的，除非customer_id 字段在两个表中可以有NULL 值，而这很可能不是这种情况。但是，如果客户可能出现在具有相同促销代码的表格的两行中，那么您应该将其更改为 COUNT(DISTINCT customer_id) ：

SELECT prom.promotion
     , prom.promo_size
     , responded.promo_responded
     , purchased.promo_purchased
     , responded.promo_responded / prom.promo_size
       AS promo_response_success_rate
FROM
    ( SELECT promotion
         , COUNT(*) AS promo_size
      FROM CUST_PROMO
      GROUP BY promotion
    ) AS prom
  LEFT JOIN 
    ( SELECT PROMO_RESPONDED AS promotion
           , COUNT(*) AS promo_responded
      FROM CUSTOMER
      GROUP BY PROMO_RESPONDED
    ) AS responded
    ON responded.promotion = prom.promotion
  LEFT JOIN
    ( SELECT PROMO_PURCHASED AS promotion
           , COUNT(*) AS promo_purchased
      FROM CUSTOMER
      GROUP BY PROMO_PURCHASED
    ) AS purchased
    ON purchased.promotion = prom.promotion

Answer 4

SELECT
  cp.promotion,
  PROMO_SIZE = COUNT(*),
  PROMO_RESPONDED = COUNT(c1.customer_id),
  PROMO_PURCHASED = COUNT(c2.customer_id),
  PROMO_RESPSUCCESSRATE = COUNT(c1.customer_id) * 100.0 / COUNT(*)
FROM CUST_PROMO cp
  LEFT JOIN CUSTOMER c1
    ON cp.customer_id = c1.customer_id AND cp.promotion = c1.PROMO_RESPONDED
  LEFT JOIN CUSTOMER c2
    ON cp.customer_id = c2.customer_id AND cp.promotion = c2.PROMO_PURCHASED
GROUP BY cp.promotion