题解 | #作答试卷得分大于过80的人的用户等级分布#

# select level, count(level) as level_cnt
# from exam_record 
# left join user_info using(uid)
# left join examination_info using(exam_id)
# where score > 80 and tag = 'SQL'
# group by level
# order by level_cnt desc



with table1 as (
    select uid, tag ,score ,level
    from exam_record left join examination_info using(exam_id)
    left join user_info using(uid)
    where tag = 'SQL' and score > 80 
)
select level , count(distinct uid) as level_cnt
from table1 

group by level 
order by level_cnt desc,level desc