从单个选择查询中提取多个数学运算输出

Question

我需要合并三个表以进行分析：活动、学生和账单。

“活跃”包含活跃学生的记录和他们活跃的科目，列：id（学生 id）int、时间（他们活跃的时间）时间戳和主题（活跃的主题）-文本

id   time                  subject
1    2020-04-23 06:53:30   Math
2    2020-05-13 09:51:22   Physics
2    2020-02-26 17:34:56   History

'Students' 是海量数据库，包含：id (student id) int, group（学生被分配到 a/b 测试的组）- text

id   group     
1    A       
2    B 
3    A
4    A

“账单”记录学生购买的课程的所有交易：id（学生 ID）int、sale_time（学生购买课程的时间）时间戳、主题（购买课程的主题）文本、金钱（支付的金额）。

id     sale_time              subject     money
1      2020-03-04 08:54:55    Math        4300
1      2020-04-08 20:43:56    Math        3200
2      2020-05-09 13:43:12    Law         8900

基本上，我们有一个学生数据库（学生），其中一些购买了课程（账单）。而一些购买的人仍然活跃（活跃）。

我需要编写一个单一的查询，我可以提取以下按它们属于 A 组还是 B 组分组：

average revenue per user: sum (money) / count (distinct Students.id)
average revenue per active user: sum (money) / count (distinct Active.id)
conversion rate (%): count (distinct Bills.id) / count (distinct Students.id)
conversion rate (active) (%): count (distinct Bills.id) / count (distinct Active.id)
conversion rate (Math) (%) (count (distinct Bills.id) where Bills.subject = Math) / (count (distinct Active.id) where Active.subject = Math)

所有这些都在一个查询中！

我用过

select sum (money)/count (distinct Students.id)
  from Students
     left join Bills using (id)
        left join Active using (id)
            group by group, Students.id

但我不知道如何在使用过滤器选择后立即进行这些数学计算。

请帮忙！

SQL 小提琴：https://www.db-fiddle.com/f/NPQR6aBf8H36XvrefJY2J/0

Answer 1

你只需要这个：

select s.[group], sum (money)/  NULLIF( count (distinct s.id),0) as 
   AvgPerUser,
   sum (money) /  NULLIF(count (distinct a.id),0) as AvgActUser,
   count (distinct b.id) /  NULLIF(count (distinct a.id),0) as CovRate,
   count (distinct b.id) /  NULLIF(count (distinct a.id),0) as ConActRate,
   (select count(distinct b2.id) from Bills as b2 where b2.subject = 'Math') /  
   NULLIF((select count ( distinct a2.id) from Active as a2  where a2.subject 
   ='Math'),0) as ConRateMath
   from Students as s
   left join Bills as b on s.id = b.id
    left join Active as a on s.id = a.id
        group by s.[group]

Answer 2

我建议在加入之前删除重复项，然后使用窗口函数：

select s.group, avg(b.money)as AvgPerUser,
       sum(b.money) /  nullif(count(a.id), 0) as AvgActUser,
       count(b.id) / nullif(count(s.id), 0) as CovRate,
       count(b.id) / nullif(count(a.id),0) as ConActRate,
       count(b.id) filter (where s.subject = 'Math') * 1.0 / count(*) filter (where s.subject = 'Math') as ConRateMath
from Students s left join
     (select b.id, sum(money) as money
      from bills b
      group by b.id
     ) b
     on s.id = b.id left join
     (select distinct a.id from active a
     ) a
     on s.id = a.id
group by s.group;  

注意：我认为您不希望 s.id 在 GROUP BY 中。那真的不会聚合任何东西。