题解 | #平均活跃天数和月活人数#

select
    concat (
        year (submit_time),
        if (
            month (submit_time) < 10,
            concat ('0', month (submit_time)),
            submit_time
        )
    ) `month`,
    round(count(DISTINCT uid,date(submit_time)) / count(DISTINCT uid), 2) avg_active_days,
    count(DISTINCT uid) mau
from
    exam_record
where
    submit_time is not null
    and year (submit_time) = 2021
group by
    month