题解 | #作答试卷得分大于过80的人的用户等级分布#

select
    t.level level,
    count(*) level_cnt
from
    (

select
    ui.uid uid,
    ui.level level
    from
        exam_record er
    left join examination_info ei on er.exam_id = ei.exam_id
    left join user_info ui on er.uid = ui.uid
    where
            ei.tag = 'SQL'
            and score > 80
    ) as t  group by t.level order by level_cnt desc