select
concat (year (submit_time), "0", month (submit_time)) month,
round(
count(distinct uid, left (submit_time, 10)) / count(distinct uid),
2
) avg_active_days,
count(distinct uid) mau
from
exam_record
where
year (submit_time) = 2021
group by
concat (year (submit_time), "0", month (submit_time))

京公网安备 11010502036488号