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)
from user_profile u
join question_practice_detail q
on u.device_id = q.device_id
join question_detail qd
on q.question_id = qd.question_id
group by university, difficult_level
having university = '山东大学'

这道题告诉我们having在group by后面 不一定要和where连用

然后Join默认就是Inner join