数据库工程师或者ETL工程师可能会遇到的问题:
查询连续登陆的天数,起始登陆日期,结束登陆日期,连续登陆天数
表结构和数据(有重复数据)如下:
原理:使用Rank 排序,得到序号,登陆日期和需要相减,得到起始日期作为附注日期,再根据用户id和辅助日期进行分组即可。
废话不多说直接上代码:
MySQL(不支持rank函数,通过白能量的形式实现):
select user_id, min(login_time) 起始登录日期, max(login_time) 结束登录日期, count(login_time) 连续登录天数 from (select *, DATE_SUB(login_time,INTERVAL rank DAY) 辅助日期列 from (select *,@rank := @rank + 1 AS rank from (select distinct user_id,DATE(login_time) as login_time from user_login_log) as a, (SELECT @rank := 0) as tmp order by user_id,login_time) as a) as c group by user_id,辅助日期列 -- having count(login_time) > 3
MySQL 参考自 :https://blog.csdn.net/github_38426094/article/details/80492093
SQL Server :
select id, min(date_time) 起始登录日期, max(date_time) 结束登录日期, count(date_time) 连续登录天数 from (select *, DATEADD(day, -od, date_time) 辅助日期列 from (select *,rank() over(partition by id order by date_time) as od from (select distinct id,date_time from testdate) as a) as b) as c group by id,辅助日期列
以上