Question

我有两个桌子。进入第一个表（活动），有：user_id，会话和login_time。在第二（付款）中，只有一列-user_id。

我需要计算用户在01/01 / 2018-01 / 05/2018期间通过“付款”进行了多少次会话。

activity

user_id   sessions   login_time
1           6         01.01.2018
2           2         01.01.2018
1           1         02.01.2018
4           3         02.01.2018
1           2         03.05.2018
3                     04.02.2018

表

payments
user_id
1
3
4

我的代码（计算错误）：

select login_time, avg(activity.sessions) AS
average_sessions
from activity inner join
payments
on payments.user_id = activity.user_id
where activity.login_time between '2018-01-01' and '2018-05-01' 
group by activity.login_time;

感谢帮助！

输出：

output
login_time    average_sessions
01.01.2018     6
02.01.2018     2
03.01.2018     0
04.01.2018     0
...

Answer 1

寻找样品
似乎您需要将会话总数除以总天数

  select  user_id, sum(sessions)/t.num_days
  from activity 
  cross join  (
  select count(distinct login_time) num_days
  from activity
  ) t 
  group by user_id

，并且如果您只需要user_id

  select  user_id, sum(sessions)/t.num_days
  from activity 
  cross join  (
  select count(distinct login_time) num_days
  from activity
  ) t 
  inner join  payments p on p.user_id = activity.user_id
  group by user_id

并期待您形成的结果似乎需要

  select t1. login_time, ifnull(t2.avg_sess) average_sessions
  from (
  select distinct login_time 
  from activity 
  ) t1  
  left join (
  select a.login_time, avg(a.session) avg_sess
  from activity a
  inner join payments p on a.user_id  = p.user_id
  group by a.login_time
  ) t2 on t1.login_time = t2.login_time

Answer 2

根据您的预期输出，使用以下内容：

SELECT login_time, 
       SUM(activity.sessions)/COUNT(DISTINCT activity.user_id) AS average_sessions
FROM activity 
INNER JOIN payments ON payments.user_id = activity.user_id
WHERE activity.login_time BETWEEN '2018-01-01' and '2018-05-01' 
GROUP BY activity.login_time;

如何计算平均值？

2 个答案: