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题解 | #统计每个用户的平均刷题数#

select university, difficult_level, count(qpd.question_id) / count(distinct qpd.device_id) from user_profile as up &nb...

2023-03-14

0 220

题解 | #21年8月份练题总数#

select did_cnt, question_cnt from (select count(distinct device_id) as did_cnt, count(question_id) as question_cnt, (case when date like "...

2023-03-13

0 227