mysql - 如何计算平均值？

Question

我有两张桌子。进入第一个表（活动）有：user_id、sessions和login_time。在第二个（付款）中只有一列 - user_id。

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

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

桌子

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
...

标签： mysqlsqldatabasejoinselect

Answer 1

查看您的样本
似乎您需要会话总和除以总天数

  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  

如果您只需要付款中的 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  

并且看着你形成的结果似乎你需要

  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