题解 | #月均完成试卷数不小于3的用户爱作答的类别#

select e2.tag , count(tag) as tag_cnt
from exam_record e1 left join examination_info e2 using(exam_id)
where e1.uid in
    (select e1.uid
     from exam_record e1 
     where e1.submit_time is not NULL
     group by e1.uid, DATE_FORMAT(submit_time, "%Y%m")
     having count(e1.submit_time) > 2
    ) 
group by e2.tag    
order by tag_cnt desc;