SELECT university,
avg(question_cnt) as avg_question_cnt,
avg(answer_cnt) as avg_answer_cnt
from user_profile
group by university
having avg_question_cnt<5 or avg_answer_cnt<20

having是这样子用的，当时整理知识点的时候不怎么明白，现在实际运用稍微明白了一点点了