梦梦冲啊

分类

/ 注册

全部文章（共10篇）

题解 | 统计每个产品的销售情况

with base as ( #先把基础表字段整理出来 select od.*, pd.unit_price, customer_age, case when cu.customer_age > 60 then...

2025-09-04

0 10

题解 | 统计商家不同会员每日访问人次及访问人数

with t as( #题目不够严谨，未考虑到访问时间和离开时间隔天的情况 select user_id, visit_time time from visit_tb union all select user_id, if(datediff(leave_time,...

2025-02-28

9 52

题解 | 查询连续登陆的用户

with t1 as( select lt.user_id, date_format(log_time,'%Y-%m-%d') log_date, dense_rank()over(partition by lt.user_id order by date_format(l...

2025-02-28

0 45

题解 | 每个商品的销售总额

select p.name product_name, sum(quantity) total_sales, rank()over(partition by p.category order by p.category,sum(quantity) desc,p.product...

2025-02-28

0 50

题解 | 推荐内容准确的用户平均评分

with t as(#先把推荐正确用户找出来，并去重，以防同一个用户多次推荐正确 select round(avg(score),3) avg_score from recommend_tb rt inner join user_action_tb uat on rt.rec_user =...

2025-02-27

0 46

题解 | 统计所有课程参加培训人次

#数，的个数来判断 select sum((length(course) - length(replace(course,',','')))+1) staff_nums from cultivate_tb

2025-02-27

4 73

题解 | 查询连续入住多晚的客户信息？

with t as( select ct.room_id, ct.user_id, gt.room_type, datediff(checkout_time,checkin_time) days from checkin_tb ct left join guestro...

2025-02-27

1 51

题解 | 分析客户逾期情况

select pay_ability, concat(round(count(overdue_days)*100/count(pay_ability),1),'%')overdue_ratio from customer_tb ct left join loan_tb lt usin...

2025-02-26

0 40

题解 | 最长连续登录天数

with t1 as( #第一步，先按date排序，注意用dense_rank可以避免出现同一天多次登陆情况 select user_id, fdate, dense_rank()over(partition by user_id order by fdate) drn fr...

2025-02-26

0 57

题解 | #博客地址修复#

select device_id, replace(replace(trim(blog_url),'http://','http:/'),'http:/','http://') blog_url from user_submit看了很多，感觉这样是最简单的操作了，先去除首尾空字符后，把所...

2022-10-21

8 249