题解 | #牛客每个人最近的登录日期(三)#

sql script

# first day login
# select user_id, min(date)
# from login
# group by user_id


# second day 
# select user_id, DATE_ADD(min(date),INTERVAL 1 DAY)
# from login
# group by user_id


# all users
# select count(distinct user_id)
# from login


# result
select round(count(distinct user_id)/(select count(distinct user_id) from login),3)
from login
where (user_id,date)
in (select user_id, DATE_ADD(min(date),INTERVAL 1 DAY)
    from login
    group by user_id)