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) tag_cnt from exam_record 
join examination_info using(exam_id) 
WHERE  uid in (select uid from (select uid,count(*)/count(distinct date_format(submit_time,'%Y%m')) month_cnt from exam_record
where submit_time is not null
group by uid
having month_cnt>=3) a)
group by tag
order by tag_cnt desc
;

认为答案有问题，首行count(tag)计数的话，会包括未交卷的数据，应该以count(submit_time)计数