手机建设网站制作,顺义企业建站费用,asp wordpress,网站备案注意#x1f345; 1、专栏介绍 「SQL面试题库」是由 不是西红柿 发起#xff0c;全员免费参与的SQL学习活动。我每天发布1道SQL面试真题#xff0c;从简单到困难#xff0c;涵盖所有SQL知识点#xff0c;我敢保证只要做完这100道题#xff0c;不仅能轻松搞定面试#xff0… 1、专栏介绍 「SQL面试题库」是由 不是西红柿 发起全员免费参与的SQL学习活动。我每天发布1道SQL面试真题从简单到困难涵盖所有SQL知识点我敢保证只要做完这100道题不仅能轻松搞定面试代码能力和工作效率也会有明显提升。 1.1 活动流程 整理题目西红柿每天无论刮风下雨保证在8am 前更新一道新鲜SQL面试真题。粉丝打卡粉丝们可在评论区写上解题思路或者直接完成SQL代码有困难的小伙伴不要着急先看别人是怎么解题的边看边学不懂就问我。交流讨论为了方便交流讨论可进入 数据仓库 。活动奖励我每天都会看评论区和群里的内容对于积极学习和热心解答问题的小伙伴红包鼓励以营造更好的学习氛围。 1.2 你的收获 增强自信搞定面试在求职中SQL是经常遇到的技能点而这些题目也多数是真实的面试题刷题可以让我们更好地备战面试增强自信提升自己的核心竞争力。 巩固SQL语法高效搞定工作通过不断练习能够熟悉SQL的语法和常用函数掌握SQL核心知识点提高SQL编写能力。代码能力提升了工作效率自然高了。 提高数据处理能力、锻炼思维能力SQL是数据处理的核心工具通过刷题可以让我们更好地理解数据处理的过程提高数据分析的效率。SQL题目的难度不一需要在一定时间内解决问题培养了我们对问题的思考能力、解决问题的能力和对时间的把控能力等。 2、今日真题 题目介绍 The Most Recent Three Orders the-most-recent-three-orders 难度中等 SQL架构 Table: Customers------------------------
| Column Name | Type |
------------------------
| customer_id | int |
| name | varchar |
------------------------
customer_id is the primary key for this table.
This table contains information about customers.Table: Orders------------------------
| Column Name | Type |
------------------------
| order_id | int |
| order_date | date |
| customer_id | int |
| cost | int |
------------------------
order_id is the primary key for this table.
This table contains information about the orders made by customer_id.
Each customer has one order per day.Write an SQL query to find the most recent 3 orders of each user. If a user ordered less than 3 orders return all of their orders. Return the result table sorted by customer_namein ascending order and in case of a tie by the customer_idin ascending order. If there still a tie, order them by the order_datein descending order. The query result format is in the following example: Customers ------------------------ | customer_id | name | ------------------------ | 1 | Winston | | 2 | Jonathan | | 3 | Annabelle | | 4 | Marwan | | 5 | Khaled | ------------------------ Orders ----------------------------------------- | order_id | order_date | customer_id | cost | ----------------------------------------- | 1 | 2020-07-31 | 1 | 30 | | 2 | 2020-07-30 | 2 | 40 | | 3 | 2020-07-31 | 3 | 70 | | 4 | 2020-07-29 | 4 | 100 | | 5 | 2020-06-10 | 1 | 1010 | | 6 | 2020-08-01 | 2 | 102 | | 7 | 2020-08-01 | 3 | 111 | | 8 | 2020-08-03 | 1 | 99 | | 9 | 2020-08-07 | 2 | 32 | | 10 | 2020-07-15 | 1 | 2 | ----------------------------------------- Result table: -------------------------------------------------- | customer_name | customer_id | order_id | order_date | -------------------------------------------------- | Annabelle | 3 | 7 | 2020-08-01 | | Annabelle | 3 | 3 | 2020-07-31 | | Jonathan | 2 | 9 | 2020-08-07 | | Jonathan | 2 | 6 | 2020-08-01 | | Jonathan | 2 | 2 | 2020-07-30 | | Marwan | 4 | 4 | 2020-07-29 | | Winston | 1 | 8 | 2020-08-03 | | Winston | 1 | 1 | 2020-07-31 | | Winston | 1 | 10 | 2020-07-15 | -------------------------------------------------- Winston has 4 orders, we discard the order of 2020-06-10 because it is the oldest order. Annabelle has only 2 orders, we return them. Jonathan has exactly 3 orders. Marwan ordered only one time. We sort the result table by customer_name in ascending order, by customer_id in ascending order and by order_date in descending order in case of a tie. Follow-up: Can you write a general solution for the most recent norders? sql
select name customer_name ,customer_id,order_id,order_date
from (
select name ,o.customer_id,order_id,order_date ,rank()over(partition by o.customer_id order by order_date desc) rk
from Orders o left join Customers c
on o.customer_idc.customer_id
)t1
where rk 3
order by customer_name ,customer_id,order_date desc已经有灵感了在评论区写下你的思路吧
