题解 | #统计每个学校各难度的用户平均刷题数#

SELECT uu university,dl difficult_level,avg(cc) avg_answer_cnt
FROM(SELECT u.university uu,u.device_id,qpd_qd.d_lev dl,count(*) cc
    FROM user_profile u
    JOIN (
        SELECT qpd.device_id d_id,qpd.question_id q_id,qd.difficult_level d_lev
        FROM question_practice_detail qpd
        JOIN question_detail qd
        ON qpd.question_id = qd.question_id
    ) qpd_qd
    ON u.device_id = qpd_qd.d_id
    GROUP BY u.device_id,qpd_qd.d_lev) u_qpd_qd
GROUP BY uu,dl;

先理清楚思路再解决问题，上来就写代码改来改去好崩溃！