题解 | #统计每个用户的平均刷题数#

select user_profile.university as university,
       question_detail.difficult_level as difficult_level,
       count(question_practice_detail.question_id)/count(distinct question_practice_detail.device_id) as avg_answer_cnt
from 
    question_practice_detail 
    INNER JOIN user_profile ON user_profile.device_id = question_practice_detail.device_id 
    INNER JOIN question_detail ON question_detail.question_id = question_practice_detail.question_id 
where university = '山东大学'
group by difficult_level

注意子句顺序，group by 子句在where子句后面