题解 | #21年8月份练题总数#

不同学校，不同程度：group by 1,2 # 分组有次序

执行顺序：先分组再count

inner join：只显示符合条件的值

SELECT
university,
difficult_level,
round(count(qp.question_id)/ count(DISTINCT qp.device_id),4) AS avg_answer_cnt
FROM user_profile ue
INNER JOIN question_practice_detail qp USING(device_id)
INNER JOIN question_detail qd USING(question_id)
GROUP BY university,difficult_level;