with cte_get_uid as (
    select uid,
    level
    from user_info
    join exam_record using(uid)
    join examination_info using(exam_id)
    where tag='SQL' and score>80
)
# with 子查询获取符合条件的用户
select level,
count(uid) as level_cnt
# 对等级分组后计算每组uid数量
from cte_get_uid
group by level
order by level_cnt desc
# 降序排序