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题解 | 分析客户逾期情况

select ct.pay_ability, CONCAT( ROUND(SUM(CASE WHEN lt.overdue_days is not null THEN 1 ELSE 0 END)/COUNT(*)*100,1),'%') as overdue_rati...

2025-06-21

1 59

题解 | 每个月Top3的周杰伦歌曲

select month, ranking, song_name, play_pv from ( select month(fdate) as month, ROW_NUMBER() over (PARTITION BY month(fdate) ORDER BY count(play_log.so...

2025-06-21

0 87

题解 | 最长连续登录天数

with tt1 as( select fdate, user_id, row_number() over (PARTITION BY user_id order by fdate) as rn FROM tb_dau ), tt2 as( select...

2025-06-21

0 41

题解 | 每个月Top3的周杰伦歌曲

select month, ranking, song_name, play_pv from ( select month(fdate) as month, ROW_NUMBER() over (PARTITION BY month(fdate) ORDER BY count(play_log.so...

2025-06-09

0 49

题解 | #计算用户的平均次日留存率#

select count(date2)/count(date1) as avg_ret from ( (SELECT device_id, date AS date1 FROM question_practice_detail GROUP BY device_id, date) as t1 le...

2024-12-08

0 111

题解 | #计算用户8月每天的练题数量#

select date_format(date,'%d') as day, count(question_id) as question_cnt from question_practice_detail where date between "2021-08-01" and ...

2024-12-07

0 68

题解 | #计算25岁以上和以下的用户数量#

(select "25岁以下" as age_cut, count(distinct device_id) as number from user_profile where age < 25 or age is null) union all (select &qu...

2024-12-07

0 95

题解 | #查找山东大学或者性别为男生的信息#

(select device_id,gender,age,gpa from user_profile where university = "山东大学" ) union all (select device_id,gender,age,gpa from user_profile ...

2024-12-07

0 65

题解 | #分组过滤练习题#

select university, avg(question_cnt) as avg_question_cnt, avg(answer_cnt) as avg_answer_cnt from user_profile group by university having avg_question...

2024-12-07

0 65