题解 | 统计每个用户的平均刷题数

# select university, difficult_level,
# round(count(answer_cnt)/count(distinct u.device_id),4) as avg_answer_cnt
# from user_profile u, question_practice_detail q, question_detail qd 
# where u.device_id = q.device_id and q.question_id = qd.question_id
# and u.university = '山东大学'
# group by 
#     u.university,
#     qd.difficult_level;



select university, difficult_level,
round(count(answer_cnt)/count(distinct u.device_id),4) as avg_answer_cnt
from user_profile u join question_practice_detail q on u.device_id = q.device_id
join question_detail qd on q.question_id = qd.question_id
group by u.university, qd.difficult_level
having u.university = '山东大学';