题解 | 统计每个用户的平均刷题数

# 关键是要搞清楚avg_answer_cnt的计算方式
select
    university,
    difficult_level,
	# 每个难度分组中，
	# question_id行数：该难度的问题被回答的总次数
	# 不同的device_id个数：回答该难度的问题的用户数
    round(count(qpd.question_id)/count(distinct qpd.device_id),4) as avg_answer_cnt
from
# 要查询的表都关联到一起，取交集
user_profile up
inner join question_practice_detail qpd
on up.device_id = qpd.device_id
inner join question_detail qd
on qpd.question_id = qd.question_id
# 范围为山东大学用户
where university = '山东大学'
# 按难度分组聚合
group by difficult_level
order by difficult_level;