select tag,count(start_time) as tag_cnt
from exam_record t1
left join examination_info t2
on t1.exam_id = t2.exam_id
where uid in
(
select uid from(
select uid,count(submit_time)/count(distinct date_format(submit_time,"%Y%m")) cnt
from exam_record
group by uid
having cnt >= 3)a
)
group by tag
order by tag_cnt desc

京公网安备 11010502036488号