题解 | #每月及截止当月的答题情况#

select
    start_month
    , count(distinct uid) mau 
    , count(distinct new_uid) month_add_uv
    , max(count(distinct new_uid)) over(order by start_month) max_month_add_uv
    , sum(count(distinct new_uid)) over(order by start_month) cum_sum_uv
from
    (
        select
        uid
        , date_format(start_time, '%Y%m') start_month
		-- 判断是否为新注册用户
        , if(start_time = min(start_time) over(partition by uid), uid, null) new_uid
    from 
        exam_record
    ) t
group by 
    start_month
;