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题解 | min(date) second_date

with t1 as( select * from( select *, lead(date,1,0) over(partition by user_id order by date) next_date, count(status) ...

2024-11-04

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题解 | #牛客的课程订单分析(四)#

with t1 as( select * from( select *, count(status) over(partition by user_id) cnt from order_info where da...

2024-11-04

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题解 | #牛客的课程订单分析(三)#分组计数筛选

select id,user_id,product_name,status,client_id,date from ( select *, count(status) over(partition by user_id) cnt from order_info ...

2024-11-04

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题解 | #牛客的课程订单分析(二)#分组计数筛选

select distinct user_id from ( select *, count(status) over(partition by user_id) cnt from order_info where date>'2025-10-15' a...

2024-11-04

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题解 | #牛客的课程订单查询：条件筛选#

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

2024-11-04

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题解 | #考试分数(五)#查询每组中位数位置上的记录

/* 1、 with t as( select *, row_number() over(partition by job order by score desc) t_rank from grade ), t2 as( select job, ...

2024-11-04

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查询各个岗位分数升序排列之后的中位数位置的范围floor()

/*1、case when语句，较为繁琐 with t as( select *, row_number() over(partition by job order by score) rk from grade ) select job, case whe...

2024-10-29

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每个岗位分数排名前2名的用户dense_rank over

select id, name, score from ( select g.id, name, score, dense_rank() over(partition by name order by...

2024-10-29

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分数大于其所在工作(job)分数的平均分的所有grade属性

/*1、并表 select id, job, score from grade left join (select job,avg(score) avg from grade group by job) t using(job) where score>avg ord...

2024-10-29

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用户刷题情况:用户的名字，以及截止到某天，累计总通过了多少题

select name u_n, date, sum(number) over(partition by user_id order by date) ps_num from login l left join passing_number p using(user_id,d...

2024-10-29

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