题解 | #月均完成试卷数不小于3的用户爱作答的类别#

select

    ei.tag tag,

    count(tag) tag_cnt

from

    exam_record er

    join examination_info ei on er.exam_id = ei.exam_id

where

    uid in (

        SELECT

            uid

        FROM

            exam_record

        WHERE

            submit_time IS NOT NULL

        GROUP BY

            DATE_FORMAT (submit_time, '%Y%m'),

            uid

        HAVING

            count(submit_time) / count(distinct DATE_FORMAT (start_time, "%Y%m")) >= 3

    )

group by

    ei.tag

order by

    tag_cnt desc;