题解 | #计算用户的平均次日留存率#

select count(b.device_id) / count(a.device_id)
from (
    select distinct device_id, date from question_practice_detail
    ) as a
left join (
    select distinct device_id, date_sub(date, interval 1 day) as date from question_practice_detail
    ) as b 
on a.date = b.date and a.device_id = b.device_id

这题做不出来 参考牛油的