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题解 | #最差是第几名(二)#

select a.grade from (select grade, sum(number) over(order by grade) t_rank, sum(number) over(order by grade desc) t_rank_desc, (select sum(number) fro...

2024-07-31

0 111

题解 | #最差是第几名(一)#

select grade, sum(number) over(order by grade) t_rank from class_grade 使用窗口函数得到累加值

2024-07-31

0 119

题解 | #实习广场投递简历分析(三)#

select a.job,a.first_year_mon,a.first_year_cnt,b.second_year_mon,b.second_year_cnt from (select job,date_format(date,'%Y-%m') first_year_mon,sum(num) ...

2024-07-31

0 136

题解 | #实习广场投递简历分析(二)#

select job,date_format(date,'%Y-%m') mon,sum(num) cnt from resume_info where year(date)<=2025 group by job,mon order by mon desc,cnt desc 使用date_fo...

2024-07-31

0 145

题解 | #实习广场投递简历分析(一)#

select job, sum(num) cnt from resume_info where date_format(date,'%Y')<=2025 group by job order by cnt desc 利用groupby和聚合函数得到结果使用date_format筛选年份

2024-07-31

0 133

题解 | #牛客的课程订单分析(七)#

select a.source,count(a.source) cnt from (select ifnull(c.name,'GroupBuy') source,user_id from order_info o left join client c on o.client_id=c.id whe...

2024-07-31

0 148

题解 | #牛客的课程订单分析(六)#

select o.id, o.is_group_buy, c.name client_name from order_info o left join client c on o.client_id=c.id where datediff(date,'2025-10-15')>0 and pr...

2024-07-29

0 151

题解 | #牛客的课程订单分析(五)#

select a.user_id, min(a.date) first_buy_date, max(a.date) second_buy_date, b.cnt from (select date, user_id,row_number() over(partition by user_id ord...

2024-07-29

0 129

题解 | #牛客的课程订单分析(四)#

select user_id, min(date) first_buy_date, count(*) cnt from order_info where datediff(date,'2025-10-15')>0 and product_name in ('C++','Java','Pytho...

2024-07-29

0 129

题解 | #牛客的课程订单分析(三)#

select o.* from order_info o where o.user_id in( select user_id from order_info where datediff(date,'2025-10-15')>0 and product_nam...

2024-07-29

0 139