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题解 | #分析客户逾期情况#

select pay_ability,concat(round(count(overdue_days)/count(*)*100,1),'%') as overdue_ratio from loan_tb lt inner join customer_tb ct on lt.customer_id=...

2024-08-14

0 88

题解 | #最长连续登录天数#

select user_id, max(consec_days) as max_consec_days from ( select user_id, count(*) as consec_days ...

2024-08-14

0 92

题解 | #每个月Top3的周杰伦歌曲#

WITH a AS ( SELECT month(fdate) AS month, t1.user_id, t1.song_id, t2.song_name...

2024-08-13

7 304