题解 | #统计每个用户的平均刷题数#

#用表连接实现

select u.university,

       qd.difficult_level,

       round(count(qpd.question_id)/count(distinct u.device_id),4) as avg_answer_cnt

from user_profile u inner join question_practice_detail qpd

on u.device_id=qpd.device_id

inner join question_detail qd

on qpd.question_id = qd.question_id

group by qd.difficult_level,u.university

having u.university='山东大学';

select * from user_profile u inner join question_practice_detail qpd

on u.device_id=qpd.device_id

inner join question_detail qd

on qpd.question_id = qd.question_id

#用where实现

select u.university,

       qd.difficult_level,

       round(count(qpd.question_id)/count(distinct u.device_id),4) as avg_answer_cnt

from question_detail qd,

     user_profile u,

     question_practice_detail qpd

where u.device_id=qpd.device_id and qpd.question_id = qd.question_id and u.university='山东大学'

group by qd.difficult_level;