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sql练习-面试前刷题

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题解 | 获得积分最多的人(二)

with total_grade as ( select user_id, sum(grade_num) as grade_num, dense_rank() over (order by sum(grade_num) desc) as rn...

2026-01-12

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题解 | 获得积分最多的人(一)

with total_grade_info as ( select user_id, sum(grade_num) as grade_num from grade_info group by user_id ...

2026-01-12

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题解 | 最差是第几名(一)

select grade, sum(number) over (order by grade) as t_rank from class_grade order by grade

2026-01-09

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题解 | 实习广场投递简历分析(二)

with resume_info_month as ( select *, substr(date,1,7) as mon from resume_info where year(date) = 2025 ...

2026-01-09

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题解 | 牛客的课程订单分析(三)

with fliter_order_info as ( select * from order_info where (date > '2025-10-15') and (status = ...

2026-01-09

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题解 | 牛客的课程订单分析(二)

with fliter_order_info as ( select * from order_info where (date > '2025-10-15') and (product_n...

2026-01-09

0 16

题解 | 牛客的课程订单分析(一)

select * from order_info where (date > '2025-10-15') and (product_name in ('C++','Java','Python')) and (status = 'completed'...

2026-01-09

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题解 | 考试分数(二)

with avg_score_job as ( select job, round(avg(score),2) as avg_score from grade group by job ) select ...

2026-01-09

0 15

题解 | 牛客每个人最近的登录日期(六)

select u.name, p_n.date, sum(number) over (partition by p_n.user_id order by p_n.date) as ps_num from passing_number as p_n left j...

2026-01-09

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题解 | 牛客每个人最近的登录日期(二)

with tem_recent_login as ( select user_id, client_id, date, row_number() over (partition by user_id order by date ...

2026-01-09

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