题解 | 统计每个学校各难度的用户平均刷题数
统计每个学校各难度的用户平均刷题数
https://www.nowcoder.com/practice/5400df085a034f88b2e17941ab338ee8
select
u.university as university,
d.difficult_level as difficult_level,
round(count(p.device_id)/count(distinct(p.device_id)),4) as avg_answer_cnt
from
user_profile u,
question_practice_detail p,
question_detail d
where
p.device_id = u.device_id
and
p.question_id = d.question_id
group by
u.university,
d.difficult_level;
虽然能猜对但依旧觉得sql语法十分诡异…
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