题解 | 每天登陆最早的用户的内容喜好

select date(log_time) as log_day,a.user_id,ut.hobby
from user_action_tb ut join (select *,
       rank() over (partition by date(log_time) order by log_time) as r
from login_tb) a on a.user_id=ut.user_id
where r = 1
order by log_day;