努力学习努力赚钱的小滈

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题解 | #分析客户逾期情况#

select ct.pay_ability, CONCAT(ROUND(COUNT( CASE WHEN lt.overdue_days is not null THEN 1 END )*100/COUNT(lt.customer_id...

2024-09-27

1 99

题解 | #最长连续登录天数#

WITH numbered_days AS ( SELECT user_id, fdate, ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY fdate) AS rn FROM ...

2024-09-27

0 87

题解 | #浙大不同难度题目的正确率#

select qd.difficult_level, count( CASE WHEN qpd.result = 'right' THEN 1 END )/count(qpd.question_id) as correct_ra...

2024-09-26

0 90

题解 | #统计每个学校各难度的用户平均刷题数#

SELECT u.university, qd.difficult_level AS difficult_level, COUNT(qpd.question_id)/COUNT(DISTINCT qpd.device_id) AS avg_answer_cnt FROM ...

2024-09-26

0 133