select university,
       difficult_level,
       round((count(qpd.question_id)/count(distinct qpd.device_id)),4) as avg_answer_cnt
		//注意:共同列必须指明来自那张表 
       from question_practice_detail qpd//qpd可以分别与up、qd进行连接
       
       inner join user_profile up 
       on up.device_id = qpd.device_id
       
       inner join question_detail qd
       on qd.question_id = qpd.question_id
       
       group by university,difficult_level;
知识点补充:mysql连接分类:内连接、外连接、全连接

内连接(inner join):等值连接、不等连接、自然连接
	等值连接:公共列值相等
    不等连接:公共列值比大小(>、<、<>())
    自然连接:无需指定公共列,公共列只出现一次
    
外连接(outer join):左外连接,右外连接