比巴卜和哪步卜

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题解 | 获得积分最多的人(一)

# 首先连接表，筛选出type是add的数据 # 针对各个人进行积分求和 # 返回最大的 select distinct b.name as name, b.grade_num as grade_num from (select a.name as name, a.grade_num as gr...

2026-01-08

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题解 | 最差是第几名(一)

select grade, sum(number) over(order by grade rows between unbounded preceding and current row) as t_rank from class_grade order by grade

2026-01-07

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题解 | 实习广场投递简历分析(三)

# 先查出2025年各个语言各个月的数据 select a.job as job, a.date1 as first_year_mon, a.num as first_year_cnt, b.date2 as second_year_mon, b.num as second_year_cnt fro...

2026-01-07

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题解 | 实习广场投递简历分析(二)

select job, substring(date,1,7) as mon, sum(num) as cnt from resume_info where year(date) = '2025' group by job,substring(date,1,7) order by mon d...

2026-01-07

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题解 | 牛客的课程订单分析(七)

select case when c.name is null then 'GroupBuy' when c.name is not null then c.name end as source, count(*) as cnt from (selec...

2026-01-07

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题解 | 牛客的课程订单分析(六)

select a.id as id, a.is_group_buy as is_group_buy, case when a.is_group_buy = 'No' then c.name when a.is_group_buy = 'Yes' then null end as cl...

2026-01-07

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题解 | 牛客的课程订单分析(五)

# 先筛选，并进行窗口函数计算：算一个日期排序，算一个计数 select distinct a.user_id as user_id, min(a.date) over(partition by user_id) as second_buy_date, max(a.d...

2026-01-07

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题解 | 牛客的课程订单分析(四)

# 先筛选，并进行窗口函数计算：算一个日期排序，算一个计数 select a.user_id as user_id, a.date as first_buy_date, a.counts as cnt from (select *, dense_ran...

2026-01-07

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题解 | 牛客的课程订单分析(三)

select a.id as id, a.user_id as user_id, a.product_name as product_name, a.status as status, a.client_id as client_id, a.date...

2026-01-07

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题解 | 考试分数(五)

# 按照语言进行分组，并排序 # 找出中位数的序号，并将序号内连接到原始表格 # 再次分组排序 select c.id as id, c.job as job, c.score as score, c.ranks as t_rank from (selec...

2026-01-06

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