题解 | #作答试卷得分大于过80的人的用户等级分布#

SELECT
    level,
    count(uid) level_cnt
from
    user_info
where
    uid in (
        SELECT
            er.uid
        FROM
            `exam_record` er
            join examination_info ei on er.exam_id = ei.exam_id
        where
            tag = 'SQL'
            and score > 80
    )
group by
    level
order by
    level_cnt desc;