题解 | #月均完成试卷数不小于3的用户爱作答的类别#

select tag,count(tag) as tag_cnt from exam_record join examination_info on exam_record.exam_id=examination_info.exam_id where uid in(select uid from exam_record where submit_time is not null group by uid having count(exam_id)/count(distinct date_format(start_time,"%Y%m")) >=3) group by tag order by tag_cnt desc