又菜又爱玩^
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题解 | #获得积分最多的人(三)#
sum 开窗 + if判断 WITH grade_new AS ( SELECT *,sum(if(gi.type ='reduce',gi.grade_num * -1,gi.grade_num)) OVER (PARTITION BY name) rk FROM user ...
Mysql
2022-02-22
0
230
题解 | #获得积分最多的人(二)#
89和90 解题方法差不多吧,有点迷惑???? WITH grade_new AS ( SELECT *, sum(gi.grade_num) OVER (PARTITION BY name) rk FROM user LEFT JOIN grade_info gi O...
Mysql
2022-02-22
0
174
题解 | #牛客的课程订单分析(七)#
可以参考一下。 WITH order_new AS ( SELECT * FROM ( SELECT *, count(*) OVER (PARTITION BY user_id) rk FROM order_info o ...
Mysql
2022-02-21
0
219
题解 | #牛客的课程订单分析(五)#
感觉有点 繁琐了 SELECT t2.user_id,t2.lagdate first_buy_date,t2.date second_buy_date,t2.rk cnt FROM ( SELECT *, row_number() OVER (PARTITION BY user_id) rk2 F...
Mysql
2022-02-18
1
179
题解 | #牛客每个人最近的登录日期(五)#
会笛卡尔乘积 select t1.day1, ifnull(round(t1.incre_user / (t1.new_user /2 ),3),0.00) p from ( select sum(case when a.tmp=1 then 1 else 0...
Mysql
2021-09-27
0
329
题解 | #出现三次以上相同积分的情况#
where 竟然可以并列判断 哈哈哈学到了 select t1.date, round(no_completed/count_completed,3) p from ( select date, count(*) count_comp...
Mysql
2021-09-24
0
302
题解 | #出现三次以上相同积分的情况#
窗口函数 永远滴神! select t1.id, t1.job, score, t_rank from ( select id, job, score, dense_rank() over (partition by job order by score desc ) t_ra...
2021-08-17
0
313
题解 | #出现三次以上相同积分的情况#
窗口 + 内连接 select t1.id, t1.job, score, t_rank from ( select id, job, score, dense_rank() over (partition by job order by score desc ) t_ran...
2021-08-17
0
261
题解 | #出现三次以上相同积分的情况#
令我没想得到的是,卡在了多字段的不同方式排序上 哈哈哈 不过最后还是解出来了 select id, name, score from ( select g.id, language_id, score, dense_rank() over (partition by...
2021-08-12
0
251
题解 | #统计salary的累计和running_total#
我还以为测试用例错了, 当我看到 结果不要排序!!! 我悟了 select e.first_name from employees e join ( select first_name, rank() over (order by first_name) as tmp fr...
2021-08-08
0
256
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