如何将来自一个 SQL 选择的聚合值除以来自另一个 SQL 选择的聚合值？答案

【问题标题】：How to divide aggregated values from one SQL select by aggregated value from another SQL select?如何将来自一个 SQL 选择的聚合值除以来自另一个 SQL 选择的聚合值？
【发布时间】：2018-11-28 17:55:32
【问题描述】：

我想计算系统与多少客户通信。在问题的背景下，在小组的背景下，在节目的背景下，在该时期的渠道背景下，按百分比计算。

公式很简单：

F = X / Y

地点：

X - number of clients communicated by system.

和

Y - number of clients in analizing sample.

我有数据库表 INTERACTIONS。

表格示例：

datetime       | Issue | Group | ProgramID | Channel | CustomerID | Outcome
---------------+-------+-------+-----------+---------+------------+--------
20181126000001 | Sale  | Loans | P-1       | SMS     | 5          | Accept  
20181126000005 | Sale  | Loans | P-1       | PUSH    | 7          | Ignore  
20181126000010 | Sale  | Loans | P-2       | SMS     | 8          | Ignore  
20181126000015 | Sale  | Loans | P-3       | PUSH    | 10         | Accept

所以 X 的选择是：

SELECT 
    ISSUE, GROUP, PROGRAMID, CHANNEL, 
    COUNT(DISTINCT CUSTOMERID) AS Customers 
FROM INTERACTIONS  
GROUP BY ISSUE, GROUP, PROGRAMID, CHANNEL 
ORDER BY 1 ASC, 2 ASC, 3 ASC, 4 ASC;

Y 的选择是：

SELECT COUNT(DISTINCT CUSTOMERID) AS Customers 
FROM INTERACTIONS;

如何结合这两个 SQL 选择并获得 F=X/Y 的报告？

我想得到类似的结果：

【问题讨论】：

标签： sql count db2 report distinct

【解决方案1】：

SELECT 
     ISSUE, GROUP, PROGRAMID, CHANNEL, 
     COUNT(DISTINCT CUSTOMERID)/
    (SELECT COUNT(DISTINCT CUSTOMERID) FROM 
     INTERACTIONS) AS Customers 
FROM INTERACTIONS  
GROUP BY ISSUE, GROUP, PROGRAMID, CHANNEL
ORDER BY 1 ASC, 2 ASC, 3 ASC, 4 ASC

只需将 Y 查询放在 X 查询中

【讨论】：