题解 | #月总刷题数和日均刷题数#

SELECT any_value(COALESCE (DATE_FORMAT(submit_time, "%Y%m"),"2021汇总")) AS submit_month,
        #COALESCE未遇到空值返回submittime，否则2021汇总
        #MySQL提供了any_value()函数来抑制ONLY_FULL_GROUP_BY值被拒绝
	COUNT(submit_time) AS month_q_cnt, 
	any_value(ROUND(COUNT(submit_time)/MAX(DAY(last_day(submit_time))),3)) AS avg_day_q_cnt
    #因为day(last_day(submit_time)运算结果还是跟submit_time同样的一串数列，只有加上avg(),min()或max运算才变成了一个数值作为分母使用
FROM practice_record
WHERE YEAR(submit_time) = 2021
GROUP BY DATE_FORMAT(submit_time, "%Y%m") WITH ROLLUP
#分组汇总