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 user_info.uid) as level_cnt
from
    user_info,
    examination_info,
    exam_record
where
    user_info.uid = exam_record.uid
    AND examination_info.exam_id = exam_record.exam_id
    AND examination_info.exam_id =  '9001'
    AND exam_record.score > 80
group by 
    level
order by
    level_cnt desc,
    level desc

本体并没有什么难点，主要难点在于需要找到user_info下的uid唯一值，然后排序顺序为计数降序和level的等级降序。