题解 | #统计每个学校各难度的用户平均刷题数#

select
 university, 
 difficult_level, 
 count(qpd.question_id)/count(distinct qpd.device_id ) as avg_answer_cnt
from 
 user_profile as up, 
 question_practice_detail as qpd , 
 question_detail as qd 
where 
 up.device_id = qpd.device_id and qpd.question_id = qd.question_id
group by up.university, qd.difficult_level

本题隐式内连接

select 字段列表 from 表1，表1，where 条件 group by 分组