题解 | #平均活跃天数和月活人数#

SELECT
    DATE_FORMAT(start_time, '%Y%m') mon,
    round(
        count(
            DISTINCT uid,
            date_format(start_time, '%Y%m%d')
        ) / count(DISTINCT uid),
        2
    ) avg_active_days,
    count(DISTINCT uid) mau
FROM
    exam_record e
WHERE
    score IS NOT NULL
    and year(start_time)=2021
GROUP BY
    DATE_FORMAT(start_time, '%Y%m')