题解 | #计算用户的平均次日留存率#

select m.m / n.n from (select count() as m from (select device_id, date from question_practice_detail group by device_id, DATE order by date) a, (select device_id, date(date + 1) as dateb from question_practice_detail group by device_id, DATE order by date) b where a.device_id = b.device_id and a.date = b.dateb) m, (select count() as n from (select device_id,DATE from question_practice_detail group by device_id,DATE order by date) n)n