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(b.question_id)/count(distinct(a.device_id)),4) as avg_answer_cnt from user_profile a
left join question_practice_detail b
on a.device_id = b.device_id
left join question_detail c 
on b.question_id = c.question_id
group by university,difficult_level
having difficult_level is not null

注意round的用法，另外每个 学校 不同难度的平均用户刷题数对学校和难度分组，后面加聚合函数having

count(b.question_id)/count(distinct(a.device_id)) 这个写法必须掌握