?? 1、專欄介紹

「SQL面試題庫(kù)」是由 不是西紅柿 發(fā)起，全員免費(fèi)參與的SQL學(xué)習(xí)活動(dòng)。我每天發(fā)布1道SQL面試真題，從簡(jiǎn)單到困難，涵蓋所有SQL知識(shí)點(diǎn)，我敢保證只要做完這100道題，不僅能輕松搞定面試，代碼能力和工作效率也會(huì)有明顯提升。

1.1 活動(dòng)流程

1. 整理題目：西紅柿每天無(wú)論刮風(fēng)下雨，保證在8am 前，更新一道新鮮SQL面試真題。
2. 粉絲打卡：粉絲們可在評(píng)論區(qū)寫上解題思路，或者直接完成SQL代碼，有困難的小伙伴不要著急，先看別人是怎么解題的，邊看邊學(xué)，不懂就問我。
3. 交流討論：為了方便交流討論，可進(jìn)入 數(shù)據(jù)倉(cāng)庫(kù) 。
4. 活動(dòng)獎(jiǎng)勵(lì)：我每天都會(huì)看評(píng)論區(qū)和群里的內(nèi)容，對(duì)于積極學(xué)習(xí)和熱心解答問題的小伙伴，紅包鼓勵(lì)，以營(yíng)造更好的學(xué)習(xí)氛圍。

1.2 你的收獲

1. 增強(qiáng)自信，搞定面試：在求職中，SQL是經(jīng)常遇到的技能點(diǎn)，而這些題目也多數(shù)是真實(shí)的面試題，刷題可以讓我們更好地備戰(zhàn)面試，增強(qiáng)自信，提升自己的核心競(jìng)爭(zhēng)力。

2. 鞏固SQL語(yǔ)法，高效搞定工作：通過不斷練習(xí)，能夠熟悉SQL的語(yǔ)法和常用函數(shù)，掌握SQL核心知識(shí)點(diǎn)，提高SQL編寫能力。代碼能力提升了，工作效率自然高了。

3. 提高數(shù)據(jù)處理能力、鍛煉思維能力：SQL是數(shù)據(jù)處理的核心工具，通過刷題可以讓我們更好地理解數(shù)據(jù)處理的過程，提高數(shù)據(jù)分析的效率。SQL題目的難度不一，需要在一定時(shí)間內(nèi)解決問題，培養(yǎng)了我們對(duì)問題的思考能力、解決問題的能力和對(duì)時(shí)間的把控能力等。

?? 2、今日真題

題目介紹： The Most Recent Three Orders the-most-recent-three-orders

難度中等

SQL架構(gòu)

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_name

in ascending order and in case of a tie by the

customer_id

in ascending order. If there still a tie, order them by the

order_date

in 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 文章來(lái)源地址http://www.zghlxwxcb.cn/news/detail-540505.html

n

orders?

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_id=c.customer_id
)t1
where rk <=3
order by customer_name ,customer_id,order_date desc

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