题解 | #统计每个用户的平均刷题数#

select university,
difficult_level,
count(qpd.question_id)/count(distinct(qpd.device_id))
from question_practice_detail as qpd 
left join user_profile as up on qpd.device_id = up.device_id
left join question_detail as qd on qpd.question_id = qd.question_id
where university = '山东大学'
group by difficult_level


# sql语句的书写顺序:
# select >> from >> where >> group by >> having >> order by >> limit
# 注意：
# 1. select和from是必须的；
# 2. where和having不能同时使用;
# 3. having和group by联合使用；

# sql语句的解析顺序 :
# from >> on>> join >> where >> group by >> having >> select >> distinct >> order by >> limit
# 注意：虽然select在having后执行，但是mysql中仍然可以在having中使用select语句定义的别名