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题解 | #统计每个用户的平均刷题数#

select u.university, t.difficult_level, count(d.question_id)/count(distinct d.device_id) from user_profile as u, question_practi...

2023-02-08

7 1963

题解 | #统计每个学校各难度的用户平均刷题数#

select university,difficult_level, count(*)/count(distinct d.device_id) from user_profile as u, question_practice_detail as d, q...

2023-02-08

0 262