如何组合SELECT语句以允许我计算SQL Server中的百分比，成功和失败？

Question

想象一个表：

CUST_PROMO（CUSTOMER_ID，促销），其被用作每一个客户收到促销之间的映射。

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

这会让我们的客户总数在每个推广。现在

，我们已经得到了客户（CUSTOMER_ID，PROMO_RESPONDED，PROMO_PURCHASED），其中列出了在客户和推广得到了客户响应，并得到了他们购买。

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 

这都是不言自明的;现在我已经掌握了每个促销成功的人数。

但是;我想最终是[以CSV格式]

PROMOTION,PROMO_SIZE,PROMO_RESPONDED,PROMO_PURCHASED,PROMO_RESPSUCCESSRATE,blah 

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

我不知道如何做到这一点。我可以联合上面的三个查询;但那实际上并不是我想要的。我想创建一个内存表，插入每个促销价值，然后做一个更新声明与一个联合反对它来设置每个值 - 但这很杂乱;并要求每个表/选择语句都有一个新的UPDATE语句。我也可以为每个结果集创建一个临时表，然后将它们结合在一起;但真的;谁想要这样做？

我想不出加入这个数据让任何意义的任何方式;因为我正在处理聚合。

所以，在最好的，我需要的是，像UNION，将结合的结果集的功能，反而会像相结合的列上的一个键，并添加这些列，而不是工会的它增加了行。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

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 

Answer 2

工作的呢？我不确定分割和乘法运算符，但是我相信我的逻辑是好的。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 

使用解码的清洁器的解决方案。仍不能确定的数学工作

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 3

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 4

是的，我认为JOIN荷兰国际集团三个集合查询是要走的路。 LEFT JOIN是为了防止某些促销没有回应或没有购买。

我也改变了COUNT(customer_id)到COUNT(*)。其结果是一样的，除非customer_id域可以有两个表这最有可能是没有的情况下NULL值。但是，如果一个客户可能会出现在同一个促销代码表的两行，那么你应该改变那些为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