我有明珠一颗

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题解 | 每个顾客购买的最新产品名称

select customer_id, customer_name, product_name as latest_order from orders join customers using(customer_id) join products using(product_id) where (c...

2025-07-18

1 63

题解 | 统计每个产品的销售情况

with prod_mon_sales as ( select product_id, unit_price, month(order_date) as mon, sum(quantity*unit_price) as mon_total_sales, sum(quantity) as mon_to...

2025-07-18

1 71

题解 | 完成员工考核试卷突出的非领导员工

with avg_s as ( select tag, avg(timestampdiff(second, start_time, submit_time)) as avg_time, avg(score) as avg_acore from exam_record join examinatio...

2025-07-17

0 71

题解 | 短视频直播间晚上11-12点之间各直播间的在线人数

select room_id, room_name, count(distinct user_id) as user_count from user_view_tb join room_info_tb using(room_id) where in_time <= '24:00:00' and...

2025-07-17

0 63

题解 | 被重复观看次数最多的3个视频

select cid, round(cast(count(*) as float), 3) as pv, row_number() over(order by count(*) desc, release_date desc) as rk from play_record_tb join cours...

2025-07-17

1 72

题解 | 请写出计算粉丝ctr的sql语句

select sum(read_num)/sum(show_num) as fans_ctr from a join b using(author_id) join c using(content_id, fans_id) 用 using 会很简洁

2025-07-17

8 75

题解 | 返回购买 prod_id 为 BR01 的产品的所有顾客的电子邮件（一）

select cust_email from Customers where cust_id in ( select cust_id from Orders where order_num in ( select order_num from OrderItems where prod_id = &...

2025-07-16

0 66

题解 | 获得积分最多的人(三)

select id, name, grade_num from ( select user_id, sum(if(type = 'reduce', grade_num * (-1), grade_num)) as grade_num, dense_rank() over(order b...

2025-07-16

1 58

题解 | #删除字符串中出现次数最少的字符#

import java.util.Scanner; import java.util.HashMap; import java.util.Collections; import java.util.Map.Entry; public class Main { public static v...

2023-11-10

0 264

题解 | #求最小公倍数#

l = list(map(int, input().strip().split())) a, b = max(l), min(l) while b > 0: temp = a % b a = b b = temp print(l[0] * l[1] // a)

2023-10-23

0 129