在MySql中查找新客户和老客户


Find new and repeat customers datewise in MySql

我有一个带有user_id和visited_at列的访问表。我想知道这个月每天有多少是新客户和老客户。

user_id      visited_at                
--------------------------------------------------------------------------
1750        2015-01-28 12:38:59
1920        2015-01-19 17:20:20
1009        2015-01-17 18:20:12
1920        2015-01-17 15:10:10
1080        2015-01-13 20:18:41
1920        2014-04-04 10:31:15
1750        2013-10-04 10:39:20

2015年1月,用户1750和1920访问了同一个地方,因此重复客户总数为2。2015年4月,用户1750、1920和1080访问了同一个地方,因此重复客户总数为3。输出应该类似于这个

October 2013
Month               New        Repeat
----------------------------------------------
2013-10-04          1          0
April 2014
Month               New        Repeat
----------------------------------------------
2014-04-04          1          0
January 2015
Month               New        Repeat
----------------------------------------------
2015-01-13          1          0
2015-01-17          1          1
2015-01-19          0          1
2015-01-28          0          1

Chopra。事实上,我确实误解了你在评论中提出的问题。这与按月的结构非常相似,只是按日期聚合:

select date(v.visited_at),
       count(case when v.visited_at = vv.minva then user_id end) as num_new_users,
       (count(distinct user_id) - count(case when v.visited_at = vv.minva then user_id end)
       ) as num_repeat_users
from visits v join
     (select user_id, min(visited_at) as minva
      from visits t
      group by user_id
     ) vv
     on v.user_id = vv.user_id
group by date(v.visited_at)
order by date(v.visited_at);