题解 | 统计每个用户的平均刷题数

select
    university,
    difficult_level,
    count(q.question_id) / count(distinct (q.device_id)) avg_answer_cnt
#from where写法
from
    user_profile u,
    question_practice_detail q,
    question_detail d
where
    u.device_id = q.device_id
    and q.question_id = d.question_id
#下方的inner join写法也对，注意inner可以省略,第一个inner join
#可以换成left outer join，第二个inner join不能换，因为inner是取共同值，第二个表连接第三个表去共同值就会缺失，第一个表和第二个不管用哪一个都不会缺失
# from
#     user_profile u  left outer join question_practice_detail q 
#     on(u.device_id = q.device_id) inner join  question_detail d
#     on (q.question_id = d.question_id)
and
    university = "山东大学"
group by
    difficult_level;