select
e2.tag,
count(*) cnt
from exam_record e1 left join examination_info e2 on e1.exam_id=e2.exam_id
WHERE
e1.uid in (
select uid from exam_record
WHERE
submit_time is not NULL
group by uid ,date_format(start_time,'%Y%m')
having count(*)>=3
)
group by e2.tag
order by cnt desc
其实,了解题意 就好做了,是求用户ID完成作答大于3情况下,查看符合条件用户ID,试卷类别,这些试卷类中有几次作答