题目：求每个学校(university)每种性别(gender)的用户数、30天内平均平均活跃天数(active_days_within_30)和平均发帖数量(question_cnt)。
我的尝试：
select count(device_id) as user_num,
round(AVG(active_days_within_30),1) as avg_active_day,
round(AVG(question_cnt),1) as avg_question_cnt,
from user_profile
where group by university and gender
错误点（暂时发现的）：
1、语法错误where和group by连用是不对的，删掉where
2、越界数组行为(虽然不知道牛客咋判定的越界，但是我没觉得我越界了（气壮！）)：看了讨论区，在select后面加了个university和gender，又出现了越界行为（喵喵喵的！）要删掉and。
敲！还要注意顺序，答案：
select
gender,
university,
count(device_id) as user_num,
round(AVG(active_days_within_30),1) as avg_active_day,
round(AVG(question_cnt),1) as avg_question_cnt,
from user_profile
group by gender,university
标签：cnt,发帖,gender,university,question,分组,SQL,active,avg From： https://www.cnblogs.com/buki233/p/17225117.html