题解 | #统计每个用户的平均刷题数#

select 
    s1.university,
    s3.difficult_level,
    round(count(s2.question_id)/count(distinct s2.device_id),4) as avg_answer_cnt
from user_profile as s1
inner join question_practice_detail as s2
on s1.device_id = s2.device_id and s1.university='山东大学'

inner join question_detail as s3
on s2.question_id = s3.question_id

group by s3.difficult_level,s1.university;


select 
    "山东大学" as university,
    difficult_level,
    count(qpd.question_id) / count(distinct qpd.device_id) as avg_answer_cnt
from question_practice_detail as qpd

inner join user_profile as up
on up.device_id=qpd.device_id and up.university="山东大学"

inner join question_detail as qd
on qd.question_id=qpd.question_id

group by difficult_level



/*
group by 后要排序n-1个字段，不排序则须如method2 将university设置成同一content
*/