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题解 | #最差是第几名(一)#

select grade, SUM(number) OVER (order by grade) as total_number from class_grade;

2023-08-30

0 219

题解 | #最差是第几名(二)#又臭又长的代码来咯

# 先找到中位数的初始位置 # 注意round是四舍五入 with t1 as ( select (case when MOD(sum(number), 2) = 0 then round(sum(number)/2, 0) else round(sum(number)/2, 0) end ...

2023-08-30

1 318

题解 | #获取有奖金的员工相关信息。#

with t1 as ( select a.emp_no, a.recevied, b.first_name, b.last_name, a.btype from emp_bonus as a left join employees as b on a.emp_no ...

2023-08-29

0 306

题解 | #牛客的课程订单分析(二)#

with t1 as ( select user_id, product_name from order_info where date > '2025-10-15' and status = 'completed' and product_name i...

2023-08-26

0 264

题解 | #获得积分最多的人(一)#

with t1 as ( select a.name, (case when b.type = 'add' then b.grade_num else -b.grade_num end ) as grade_change from grade_info as b le...

2023-08-26

0 318

题解 | #获得积分最多的人(二)#

# 运用左连接将每一个用户的积分变化记录查找出来 with t1 as ( select user_id as id, b.name, (case when a.type = 'add' then grade_num else -grade_num end) as jifen_cha...

2023-08-23

0 356

题解 | #网易云音乐推荐(网易校招笔试真题)#

# 找到user_id = 1关注的用户 with t1 as ( select follower_id from follow where user_id = 1 ), # 找到他关注的用户喜欢的音乐 t2 as ( select t1.follower_id, a.music_id from t...

2023-08-23

0 310

题解 | #考试分数(五)#

with t1 as ( select job, (case when MOD(count(score), 2) = 1 then CONVERT(count(score)/2, SIGNED INTEGER) else CONVERT(count(score)/2, SIGNED...

2023-08-21

0 298

题解 | #实习广场投递简历分析(三)#

# 找到2025年每月的数据 with t1 as ( select id, job, date, num, DATE_FORMAT(date, '%Y-%m') as first_year_mon from resume_info where date between '2...

2023-08-20

0 421

题解 | #获得积分最多的人(三)#

with t1 as ( select user_id, (case when type = 'add' then grade_num else -grade_num end) as grade_change from grade_info ), # 将每一个人的总积...

2023-08-20

1 283