SELECT u.university,SUM(e.ans_num)/COUNT(*) FROM user_profile u JOIN ( SELECT device_id,COUNT(*) AS ans_num FROM question_practice_detail GROUP BY device_id) e ON e.device_id= u.device_id GROUP BY university ORDER BY university ASC; 子查询加多表查询,子查询建立新表将device_id和该id下的答题数目作为新表的表头作为答题数目,将...