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题解 | #获得积分最多的人(三)# 来自专栏

select user.id, name, total_num from user left outer join (select user_id, sum(case type when 'add' then grade_num else -1*grade_num end) as ...

Mysql

2021-11-15

0 388

题解 | #获得积分最多的人(二)# 来自专栏

这一道题就是接着上一道题的思路继续延伸。主要是多了一步将lomit 0,1 所代表的最大值提取出来的过程。 select id, name, total_num from user left outer join (select user_id, sum(grade_num) ...

Mysql

2021-11-15

0 334

题解 | #获得积分最多的人(一)# 来自专栏

select name, max(total_num) as grade_sum from user left outer join (select user_id, sum(grade_num) as total_num from grade_info group b...

Mysql

2021-11-15

0 325

题解 | #最差是第几名(二)# 来自专栏

/* select sum(number)/2 as mid_num from class_grade; select grade, sum(number) over(order by grade) as t_rank from class_grade; select grade, sum(nu...

Mysql

2021-11-15

0 403

题解 | #最差是第几名(一)# 来自专栏

select grade, sum(number) over(order by grade) as t_rank from class_grade order by grade; sum(a) over(order by b) 窗口函数的含义是按照b列的升序顺序依次对a列进行求和。其实，最后一...

Mysql

2021-11-15

1 293

题解 | #实习广场投递简历分析(三)# 来自专栏

select a.job, a.mon as first_year_mon, a.cnt as first_year_cnt, b.mon as second_year_mon, b.cnt as second_year_cnt from (select job, left(date,7...

Mysql

2021-11-15

0 360

题解 | #实习广场投递简历分析(二)# 来自专栏

就是一些常用函数的应用，熟练掌握就好。 select job, left(date,7) as mon, sum(num) as cnt from resume_info where year(date) = 2025 group by job, mon order by mon desc, cnt...

Mysql

2021-11-15

0 233

题解 | #实习广场投递简历分析(一)# 来自专栏

/*解法1 select job, sum(num) as cnt from resume_info where date >='2025-01-01' and date <= '2025-12-31' group by job order by cnt desc; */ /*解法2 s...

Mysql

2021-11-15

0 324

题解 | #考试分数(三)# 来自专栏

/* select language_id, score, DENSE_RANK() over (order by score) as s from grade group by language_id; */ select score_rank.id, name, score_rank.score...

Mysql

2021-10-13

0 437

题解 | #牛客每个人最近的登录日期(四)# 来自专栏

-- 第一步：新用户临时表 select user_id, min(date) as m_date from login group by user_id; -- 第二步：根据日期，统计人数，得出有新用户登录时的日期及其数量 select m_date, count(distinct use...

Mysql

2021-10-10

0 418