题解 | #得分不小于平均分的最低分#

union all 的使用

SELECT DATE_FORMAT(submit_time, "%Y%m")AS submit_month,
    COUNT(submit_time) AS month_q_cnt,
    ROUND(COUNT(submit_time)/MAX(DAY(last_day(submit_time))),3) AS avg_day_q_cnt
FROM practice_record
WHERE YEAR(submit_time) = 2021
GROUP BY DATE_FORMAT(submit_time, "%Y%m")
union all
SELECT "2021汇总"AS submit_month,
    COUNT(submit_time) AS month_q_cnt,
    ROUND(COUNT(*)/31,3) AS avg_day_q_cnt
FROM practice_record
WHERE YEAR(submit_time) = 2021
order by submit_month;