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(qpd.question_id) / count(distinct qpd.device_id),
    4
  ) as avg_answer_cnt
from
  question_practice_detail as qpd
  left join user_profile as up on up.device_id = qpd.device_id
  left join question_detail as qd on qd.question_id = qpd.question_id
group by
  university,
  difficult_level

注意left join的使用，主要根据基表进行连接。因为题目中说“答题”的人，因此使用left join时需要以用户答题“题库练习明细表：question_practice_detail”为基表进行连接，即该表表示哪个用户答哪道题。