题解 | #连续两次作答试卷的最大时间窗#

select 
    uid,
    max(datediff(next_time,start_time))+1 as days_window,
    round(count(start_time)/(datediff(max(start_time),min(start_time))+1)*(max(datediff(next_time,start_time))+1),2)as avg_exam_cnt
from(
    select 
        uid,
        start_time,
        lead(start_time,1) over(partition by uid order by start_time) as next_time
    from exam_record
    where year(start_time) = '2021'
    )a
group by uid
having count(distinct date(start_time)) > 1