将识别一周内返回的活跃用户的查询

问题描述

编写一个查询来识别返回的活跃用户。回访活跃用户是指在第一次购买后的 7 天内进行了第二次购买的用户。

id  u_id item   created_at  revenue
1   109 milk    3/3/2020 0:00   123
2   139 biscuit 3/18/2020 0:00  421
3   120 milk    3/18/2020 0:00  176
4   108 banana  3/18/2020 0:00  862
5   130 milk    3/28/2020 0:00  333
6   103 bread   3/29/2020 0:00  862
7   122 banana  3/7/2020 0:00   952
8   125 bread   3/13/2020 0:00  317
9   139 bread   3/23/2020 0:00  929
10  141 banana  3/17/2020 0:00  812
11  116 bread   3/31/2020 0:00  226
12  128 bread   3/4/2020 0:00   112
13  146 biscuit 3/4/2020 0:00   362
14  119 banana  3/28/2020 0:00  127

解决方法

您可以使用窗口函数获取最早的创建日期，然后在一周内查找其他记录：

select distinct u_id
from (select t.*,min(created_at) over (partition by u_id) as min_created_at
      from t
     ) t
where created_at > min_created_at and
      created_at < min_created_at + interval 7 day;

如果您只检查客户第一次购买，并在接下来的 7 天内第二次访问，您将在第二次访问后放弃第三次购买。只需在 7 天的间隔内全局检查两次购买，如下所示：

create table t(id integer,u_id integer,item varchar(100),created_at date,revenue float);
    
 insert into t 
 values (1,109,"milk",STR_TO_DATE("3/3/2020",'%m/%d/%Y'),123),(2,139,"biscuit",STR_TO_DATE("3/18/2020",421),(3,120,176),(4,108,"banana",862),(5,130,STR_TO_DATE("3/28/2020",333),(6,103,"bread",STR_TO_DATE("3/29/2020",(7,122,STR_TO_DATE("3/7/2020",952),(8,125,STR_TO_DATE("3/13/2020",317),(9,STR_TO_DATE("3/23/2020",929),(10,141,STR_TO_DATE("3/17/2020",812),(11,116,STR_TO_DATE("3/31/2020",226),(12,128,STR_TO_DATE("3/4/2020",112),(13,146,362),(14,119,127);

select * from t as t1 where exists (select * from t as t2 where t1.u_id = t2.u_id and t1.created_at - t2.created_at > 0 and  t1.created_at - t2.created_at <= 7  );

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