题解 | #统计每个用户的平均刷题数#

  u.university,
  qd.difficult_level,
  count(qpd.question_id) / count(distinct qpd.device_id) as avg_answer_cnt
  //注意 有时候通过不了是因为用了中文的括号...
from
  question_practice_detail as qpd
  inner join user_profile as u on u.device_id = qpd.device_id
  inner join question_detail as qd on qd.question_id = qpd.question_id
where
  u.university = '山东大学'
group by
  qd.difficult_level//因为最后没有按山东大学分组，而是按了难度分类，如果按了山东大学分类则只会出现一个山东大学了