题解 | #统计每个学校各难度的用户平均刷题数#

select up.university,qd.difficult_level,
round(count(qpd.question_id)/count(distinct qpd.device_id),4) as avg_answer_cnt
from question_practice_detail as qpd
left join user_profile as up
on up.device_id=qpd.device_id
left join question_detail as qd
on qd.question_id=qpd.question_id
group by university,difficult_level

注意要将question_practice_detail作为左表，如果将user_profile作为左表的话，会出现如下的情况，这是因为user_profile中的device_id没有全部答题