最爱青草蛋糕

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题解 | 查询下订单用户访问次数？

select user_id, count(user_id) as visit_nums from ( select distinct o.user_id as user_id,date(order_time),visit_time,leave_time from order_tb as o joi...

2025-08-05

0 28

题解 | 统计各等级会员用户下订单总额

select vip,sum(price) as order_total from ( select u.user_id, vip, ifnull(order_price,0) as price from uservip_tb as u left join ( select user_id,ord...

2025-08-05

0 28

题解 | 查询连续登陆的用户

select user_id from ( select user_id,day_divided,count(*) as numbers from ( select user_id, date_sub(date(log_time),interval rank()over(partition by u...

2025-08-04

0 26

题解 | 每个商品的销售总额

select name as product_name,total_sales,category_rank from ( select p1.product_id,name,category,total_sales, rank()over(partition by category order by...

2025-08-01

0 30

题解 | 推荐内容准确的用户平均评分

select avg(score) as avg_score from ( select user_id,hobby_l,score,rec_info_l from user_action_tb as u join ( select distinct rec_user,rec_info_l from...

2025-08-01

0 30

题解 | 查询培训指定课程的员工信息

select staff_id,staff_name from ( select s.staff_id,staff_name,course from staff_tb as s join ( select staff_id,course from cultivate_tb ) as c on s....

2025-08-01

0 25

题解 | 统计所有课程参加培训人次

select sum(num) as staff_nums from ( select case when length(course) = 7 then 1 when length(course) = 15 then 2 when length(course) = 23 then 3 else ...

2025-07-31

0 17

题解 | 查询连续入住多晚的客户信息？

select user_id,g.room_id,room_type, date(checkout_time)-date(checkin_time) as days from guestroom_tb as g join ( select room_id,user_id,checkin_time,c...

2025-07-31

1 36

题解 | 获取指定客户每月的消费额

select time,sum(t_amount) as total from ( select date_format(t_time,'%Y-%m') as time,t_cus,t_type,t_amount,c_name from trade as t join ( select c_id,...

2025-07-31

0 29

题解 | 分析客户逾期情况

select pay_ability, concat(round(sum(count_number)/sum(count_1)*100,1),'%') as overdue_ratio from ( select pay_ability, case when overdue_days>0 t...

2025-07-30

0 30