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题解 | 返回购买价格为 10 美元或以上产品的顾客列表

SELECT DISTINCT cust_id FROM Orders WHERE order_num IN ( SELECT order_num FROM OrderItems WHERE item_price >= 10 ) 对cust_id去重单价＞=10元的...

2025-12-04

0 32

题解 | 今天的刷题量(一)

select name,count(*) cnt from submission s join subject su on su.id=s.subject_id where s.create_time = CURDATE() group by name,su.id order by cnt desc...

2025-12-04

0 25

题解 | 牛客的课程订单分析(六)

SELECT T3.id, T3.is_group_buy, CASE WHEN T3.is_g+ id, is_group_buy,client_id, COUNT(*) OVER (PARTITION BY user_id) ...

2025-12-02

0 21

题解 | 获取指定客户每月的消费额

SELECT DATE_FORMAT(t.t_time, '%Y-%m') AS time,--只保留年月 ROUND(SUM(t.t_amount),1) AS total FROM trade t JOIN customer c ON t.t_cus = c.c_...

2025-11-30

0 30

题解 | 牛客的课程订单分析(五)

select user_id, min(date) as first_buy_date, max(date) as second_buy_date, cnt from ( select user_id, date, row_number() over(part...

2025-11-30

0 31

题解 | 牛客的课程订单分析(四)

SELECT user_id, MIN(date) AS first_buy_date,--2025-10-15之后最小的日期 COUNT(*) AS cnt--总行数 FROM order_info WHERE status = 'completed' AND pr...

2025-11-30

0 32

题解 | 牛客的课程订单分析(二)

SELECT user_id FROM ( SELECT user_id, COUNT(*) AS COT FROM order_info WHERE date > '2025-10-15' AND status = 'completed' ...

2025-11-29

0 16

题解 | 牛客的课程订单分析(一)

select * from order_info where date>"2025-10-15"AND status="completed"and product_name IN ("C++","Java",&qu...

2025-11-29

0 29

题解 | 分析客户逾期情况

select c.pay_ability, concat(round(sum(case when overdue_days is not NULL then 1 else 0 end)/count(*)*100,1),"%") overdue_ratio from loan_tb...

2025-11-29

0 29

题解 | 考试分数(四)

SELECT job, ceil((count(*))/2) AS "start", ceil((count(*) + 1)/2) AS "end" FROM grade GROUP BY job ORDER BY job; cei...

2025-11-28

0 29