题解 | #平均活跃天数和月活人数#

SELECT
    date_format(submit_time,'%Y%m') month,
    round(count(DISTINCT concat(uid,date_format(submit_time,'%Y-%m-%d')))/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 date_format(submit_time,'%Y%m');