题解 | #作答试卷得分大于过80的人的用户等级分布#

select
    level,
    count(distinct u.uid) as level_cnt
from
    exam_record as er
    left join examination_info as ei on er.exam_id = ei.exam_id
    left join user_info as u on er.uid = u.uid
where
    u.uid in (
        select
            u.uid
        from
            exam_record as er
            left join examination_info as ei on er.exam_id = ei.exam_id
            left join user_info as u on er.uid = u.uid
        where score > 80
    )
and tag = 'SQL'
group by level
order by level_cnt desc, level desc