题解 | #统计每个学校各难度的用户平均刷题数#

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

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

多表连接：

A left join B on ...

left join C on...