力扣SQL仅数据库(1068~1084)

2024-09-03 09:36

本文主要是介绍力扣SQL仅数据库(1068~1084),希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!

1068. 产品销售分析1

需求

编写解决方案,以获取 Sales 表中所有 sale_id 对应的 product_name 以及该产品的所有 year 和 price 。
输入:
Sales 表:
+---------+------------+------+----------+-------+
| sale_id | product_id | year | quantity | price |
+---------+------------+------+----------+-------+ 
| 1       | 100        | 2008 | 10       | 5000  |
| 2       | 100        | 2009 | 12       | 5000  |
| 7       | 200        | 2011 | 15       | 9000  |
+---------+------------+------+----------+-------+
Product 表:
+------------+--------------+
| product_id | product_name |
+------------+--------------+
| 100        | Nokia        |
| 200        | Apple        |
| 300        | Samsung      |
+------------+--------------+
输出:
+--------------+-------+-------+
| product_name | year  | price |
+--------------+-------+-------+
| Nokia        | 2008  | 5000  |
| Nokia        | 2009  | 5000  |
| Apple        | 2011  | 9000  |
+--------------+-------+-------+

数据准备

Create table If Not Exists Sales (sale_id int, product_id int, year int, quantity int, price int)
Create table If Not Exists Product (product_id int, product_name varchar(10))
Truncate table Sales
insert into Sales (sale_id, product_id, year, quantity, price) values ('1', '100', '2008', '10', '5000')
insert into Sales (sale_id, product_id, year, quantity, price) values ('2', '100', '2009', '12', '5000')
insert into Sales (sale_id, product_id, year, quantity, price) values ('7', '200', '2011', '15', '9000')
Truncate table Product
insert into Product (product_id, product_name) values ('100', 'Nokia')
insert into Product (product_id, product_name) values ('200', 'Apple')
insert into Product (product_id, product_name) values ('300', 'Samsung')

代码实现

select product_name,year,price from sales s join product p on s.product_id=p.product_id;

1069. 产品销售分析2

需求

编写解决方案,统计每个产品的销售总量。
返回结果表 无顺序要求 。
输入:
Sales 表:
+---------+------------+------+----------+-------+
| sale_id | product_id | year | quantity | price |
+---------+------------+------+----------+-------+ 
| 1       | 100        | 2008 | 10       | 5000  |
| 2       | 100        | 2009 | 12       | 5000  |
| 7       | 200        | 2011 | 15       | 9000  |
+---------+------------+------+----------+-------+
Product 表:
+------------+--------------+
| product_id | product_name |
+------------+--------------+
| 100        | Nokia        |
| 200        | Apple        |
| 300        | Samsung      |
+------------+--------------+
输出:
+--------------+----------------+
| product_id   | total_quantity |
+--------------+----------------+
| 100          | 22             |
| 200          | 15             |
+--------------+----------------+

数据准备

Create table If Not Exists Sales (sale_id int, product_id int, year int, quantity int, price int)
Create table If Not Exists Product (product_id int, product_name varchar(10))
Truncate table Sales
insert into Sales (sale_id, product_id, year, quantity, price) values ('1', '100', '2008', '10', '5000')
insert into Sales (sale_id, product_id, year, quantity, price) values ('2', '100', '2009', '12', '5000')
insert into Sales (sale_id, product_id, year, quantity, price) values ('7', '200', '2011', '15', '9000')
Truncate table Product
insert into Product (product_id, product_name) values ('100', 'Nokia')
insert into Product (product_id, product_name) values ('200', 'Apple')
insert into Product (product_id, product_name) values ('300', 'Samsung')

代码实现

select product_id,sum(quantity) total_quantity from sales group by product_id;

1070. 产品销售分析3

需求

编写解决方案,选出每个售出过的产品 第一年 销售的 产品 id、年份、数量 和 价格。
结果表中的条目可以按 任意顺序 排列。
结果格式如下例所示:
示例 1:输入:
Sales 表:
+---------+------------+------+----------+-------+
| sale_id | product_id | year | quantity | price |
+---------+------------+------+----------+-------+ 
| 1       | 100        | 2008 | 10       | 5000  |
| 2       | 100        | 2009 | 12       | 5000  |
| 7       | 200        | 2011 | 15       | 9000  |
+---------+------------+------+----------+-------+
Product 表:
+------------+--------------+
| product_id | product_name |
+------------+--------------+
| 100        | Nokia        |
| 200        | Apple        |
| 300        | Samsung      |
+------------+--------------+
输出:
+------------+------------+----------+-------+
| product_id | first_year | quantity | price |
+------------+------------+----------+-------+ 
| 100        | 2008       | 10       | 5000  |
| 200        | 2011       | 15       | 9000  |
+------------+------------+----------+-------+

数据准备

Create table If Not Exists Sales (sale_id int, product_id int, year int, quantity int, price int)
Create table If Not Exists Product (product_id int, product_name varchar(10))
Truncate table Sales
insert into Sales (sale_id, product_id, year, quantity, price) values ('1', '100', '2008', '10', '5000')
insert into Sales (sale_id, product_id, year, quantity, price) values ('2', '100', '2009', '12', '5000')
insert into Sales (sale_id, product_id, year, quantity, price) values ('7', '200', '2011', '15', '9000')
Truncate table Product
insert into Product (product_id, product_name) values ('100', 'Nokia')
insert into Product (product_id, product_name) values ('200', 'Apple')
insert into Product (product_id, product_name) values ('300', 'Samsung')

代码实现

with t1 as (select *,rank() over(partition by product_id order by year)ran from sales)
select product_id,year first_year,quantity,price from t1 where ran=1;

1075 项目员工 1

需求

请写一个 SQL 语句,查询每一个项目中员工的 平均 工作年限,精确到小数点后两位。
以 任意 顺序返回结果表。
查询结果的格式如下。示例 1:
输入:
Project 表:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+Employee 表:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   | 1                |
| 4           | Doe    | 2                |
+-------------+--------+------------------+输出:
+-------------+---------------+
| project_id  | average_years |
+-------------+---------------+
| 1           | 2.00          |
| 2           | 2.50          |
+-------------+---------------+
解释:第一个项目中,员工的平均工作年限是 (3 + 2 + 1) / 3 = 2.00;第二个项目中,员工的平均工作年限是 (3 + 2) / 2 = 2.50

数据准备

Create table If Not Exists Project (project_id int, employee_id int)
Create table If Not Exists Employee (employee_id int, name varchar(10), experience_years int)
Truncate table Project
insert into Project (project_id, employee_id) values ('1', '1')
insert into Project (project_id, employee_id) values ('1', '2')
insert into Project (project_id, employee_id) values ('1', '3')
insert into Project (project_id, employee_id) values ('2', '1')
insert into Project (project_id, employee_id) values ('2', '4')
Truncate table Employee
insert into Employee (employee_id, name, experience_years) values ('1', 'Khaled', '3')
insert into Employee (employee_id, name, experience_years) values ('2', 'Ali', '2')
insert into Employee (employee_id, name, experience_years) values ('3', 'John', '1')
insert into Employee (employee_id, name, experience_years) values ('4', 'Doe', '2')

代码实现

select project_id,round(avg(experience_years),2) 
from employee e join project p on e.employee_id=p.employee_id 
group by project_id;

1076 项目员工 2

需求

编写一个解决方案来报告所有拥有最多员工的 项目。
以 任意顺序 返回结果表。
返回结果格式如下所示。示例 1:
输入:
Project table:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+
Employee table:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   | 1                |
| 4           | Doe    | 2                |
+-------------+--------+------------------+
输出:
+-------------+
| project_id  |
+-------------+
| 1           |
+-------------+
解释:
第一个项目有3名员工,第二个项目有2名员工。

数据准备

Create table If Not Exists Project (project_id int, employee_id int)
Create table If Not Exists Employee (employee_id int, name varchar(10), experience_years int)
Truncate table Project
insert into Project (project_id, employee_id) values ('1', '1')
insert into Project (project_id, employee_id) values ('1', '2')
insert into Project (project_id, employee_id) values ('1', '3')
insert into Project (project_id, employee_id) values ('2', '1')
insert into Project (project_id, employee_id) values ('2', '4')
Truncate table Employee
insert into Employee (employee_id, name, experience_years) values ('1', 'Khaled', '3')
insert into Employee (employee_id, name, experience_years) values ('2', 'Ali', '2')
insert into Employee (employee_id, name, experience_years) values ('3', 'John', '1')
insert into Employee (employee_id, name, experience_years) values ('4', 'Doe', '2')

代码实现

取最大值有两种方法:

  1. 使用子查询得出最大值

  2. 使用排序+limit 1 (有并列输出的情况不适用)

with t1 as (select project_id,count(distinct e.employee_id)con from employee e join project p on e.employee_id=p.employee_id group by project_id)
select project_id from t1 where con >= (select max(con) from t1);

1077 项目员工 3

需求

编写解决方案,报告在每一个项目中 经验最丰富 的雇员是谁。如果出现经验年数相同的情况,请报告所有具有最大经验年数的员工。
返回结果表 无顺序要求 。
结果格式如下示例所示。示例 1:
输入:
Project 表:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+Employee 表:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   | 3                |
| 4           | Doe    | 2                |
+-------------+--------+------------------+
输出:
+-------------+---------------+
| project_id  | employee_id   |
+-------------+---------------+
| 1           | 1             |
| 1           | 3             |
| 2           | 1             |
+-------------+---------------+
解释:employee_id 为 1 和 3 的员工在 project_id 为 1 的项目中拥有最丰富的经验。在 project_id 为 2 的项目中,employee_id 为 1 的员工拥有最丰富的经验。

数据准备

Create table If Not Exists Project (project_id int, employee_id int)
Create table If Not Exists Employee (employee_id int, name varchar(10), experience_years int)
Truncate table Project
insert into Project (project_id, employee_id) values ('1', '1')
insert into Project (project_id, employee_id) values ('1', '2')
insert into Project (project_id, employee_id) values ('1', '3')
insert into Project (project_id, employee_id) values ('2', '1')
insert into Project (project_id, employee_id) values ('2', '4')
Truncate table Employee
insert into Employee (employee_id, name, experience_years) values ('1', 'Khaled', '3')
insert into Employee (employee_id, name, experience_years) values ('2', 'Ali', '2')
insert into Employee (employee_id, name, experience_years) values ('3', 'John', '3')
insert into Employee (employee_id, name, experience_years) values ('4', 'Doe', '2')

代码实现

with t1 as (select *,rank() over(order by experience_years desc) ran from employee)
select distinct p.employee_id from t1 join project p on t1.employee_id=p.employee_id where ran =1;

1082 销售分析1

需求

编写解决方案,找出总销售额最高的销售者,如果有并列的,就都展示出来。
以 任意顺序 返回结果表。
返回结果格式如下所示。
示例 1:输入:
Product 表:
+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1          | S8           | 1000       |
| 2          | G4           | 800        |
| 3          | iPhone       | 1400       |
+------------+--------------+------------+
Sales 表:
+-----------+------------+----------+------------+----------+-------+
| seller_id | product_id | buyer_id | sale_date  | quantity | price |
+-----------+------------+----------+------------+----------+-------+
| 1         | 1          | 1        | 2019-01-21 | 2        | 2000  |
| 1         | 2          | 2        | 2019-02-17 | 1        | 800   |
| 2         | 2          | 3        | 2019-06-02 | 1        | 800   |
| 3         | 3          | 4        | 2019-05-13 | 2        | 2800  |
+-----------+------------+----------+------------+----------+-------+
输出:
+-------------+
| seller_id   |
+-------------+
| 1           |
| 3           |
+-------------+
解释:Id 为 1 和 3 的销售者,销售总金额都为最高的 2800。

数据准备

Create table If Not Exists Product (product_id int, product_name varchar(10), unit_price int)
Create table If Not Exists Sales (seller_id int, product_id int, buyer_id int, sale_date date, quantity int, price int)
Truncate table Product
insert into Product (product_id, product_name, unit_price) values ('1', 'S8', '1000')
insert into Product (product_id, product_name, unit_price) values ('2', 'G4', '800')
insert into Product (product_id, product_name, unit_price) values ('3', 'iPhone', '1400')
Truncate table Sales
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('1', '1', '1', '2019-01-21', '2', '2000')
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('1', '2', '2', '2019-02-17', '1', '800')
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('2', '2', '3', '2019-06-02', '1', '800')
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('3', '3', '4', '2019-05-13', '2', '2800')

代码实现

with t1 as (select seller_id,sum(price)suu from sales group by seller_id )
,t2 as (select seller_id,rank() over(order by suu desc) ran from t1)
select seller_id from t2 where ran=1;

1083. 销售分析2

需求

编写一个解决方案,报告那些买了 S8 而没有买 iPhone 的 买家。注意,S8 和 iPhone 是 Product 表中显示的产品。
以 任意顺序 返回结果表。
结果格式如下所示。
示例 1:输入:
Product table:
+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1          | S8           | 1000       |
| 2          | G4           | 800        |
| 3          | iPhone       | 1400       |
+------------+--------------+------------+
Sales table:
+-----------+------------+----------+------------+----------+-------+
| seller_id | product_id | buyer_id | sale_date  | quantity | price |
+-----------+------------+----------+------------+----------+-------+
| 1         | 1          | 1        | 2019-01-21 | 2        | 2000  |
| 1         | 2          | 2        | 2019-02-17 | 1        | 800   |
| 2         | 1          | 3        | 2019-06-02 | 1        | 800   |
| 3         | 3          | 3        | 2019-05-13 | 2        | 2800  |
+-----------+------------+----------+------------+----------+-------+
输出:
+-------------+
| buyer_id    |
+-------------+
| 1           |
+-------------+
解释:
id 为 1 的买家购买了一部 S8,但是却没有购买 iPhone,而 id 为 3 的买家却同时购买了这 2 部手机。

数据准备

Create table If Not Exists Product (product_id int, product_name varchar(10), unit_price int)
Create table If Not Exists Sales (seller_id int, product_id int, buyer_id int, sale_date date, quantity int, price int)
Truncate table Product
insert into Product (product_id, product_name, unit_price) values ('1', 'S8', '1000')
insert into Product (product_id, product_name, unit_price) values ('2', 'G4', '800')
insert into Product (product_id, product_name, unit_price) values ('3', 'iPhone', '1400')
Truncate table Sales
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('1', '1', '1', '2019-01-21', '2', '2000')
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('1', '2', '2', '2019-02-17', '1', '800')
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('2', '1', '3', '2019-06-02', '1', '800')
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('3', '3', '3', '2019-05-13', '2', '2800')

代码实现

with t1 as (select s.product_id,buyer_id,p.product_name from sales s join product p on s.product_id=p.product_id)
select distinct buyer_id from t1 where product_name='S8' andbuyer_id not in (select buyer_id from t1 where product_name ='iPhone');

1084. 销售分析3

需求

编写解决方案,报告 2019年春季 才售出的产品。即 仅 在 2019-01-01 (含)至 2019-03-31 (含)之间出售的商品。
以 任意顺序 返回结果表。
结果格式如下所示。
示例 1:输入:
Product table:
+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1          | S8           | 1000       |
| 2          | G4           | 800        |
| 3          | iPhone       | 1400       |
+------------+--------------+------------+
Sales table:
+-----------+------------+----------+------------+----------+-------+
| seller_id | product_id | buyer_id | sale_date  | quantity | price |
+-----------+------------+----------+------------+----------+-------+
| 1         | 1          | 1        | 2019-01-21 | 2        | 2000  |
| 1         | 2          | 2        | 2019-02-17 | 1        | 800   |
| 2         | 2          | 3        | 2019-06-02 | 1        | 800   |
| 3         | 3          | 4        | 2019-05-13 | 2        | 2800  |
+-----------+------------+----------+------------+----------+-------+
输出:
+-------------+--------------+
| product_id  | product_name |
+-------------+--------------+
| 1           | S8           |
+-------------+--------------+
解释:
id 为 1 的产品仅在 2019 年春季销售。
id 为 2 的产品在 2019 年春季销售,但也在 2019 年春季之后销售。
id 为 3 的产品在 2019 年春季之后销售。
我们只返回 id 为 1 的产品,因为它是 2019 年春季才销售的产品。

数据准备

Create table If Not Exists Product (product_id int, product_name varchar(10), unit_price int)
Create table If Not Exists Sales (seller_id int, product_id int, buyer_id int, sale_date date, quantity int, price int)
Truncate table Product
insert into Product (product_id, product_name, unit_price) values ('1', 'S8', '1000')
insert into Product (product_id, product_name, unit_price) values ('2', 'G4', '800')
insert into Product (product_id, product_name, unit_price) values ('3', 'iPhone', '1400')
Truncate table Sales
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('1', '1', '1', '2019-01-21', '2', '2000')
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('1', '2', '2', '2019-02-17', '1', '800')
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('2', '2', '3', '2019-06-02', '1', '800')
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('3', '3', '4', '2019-05-13', '2', '2800')

代码实现

select product_id,product_name from product where
product_id in (select product_id from sales where sale_date between '2019-01-01' and '2019-03-31') and
product_id not in (select product_id from sales where sale_date<'2019-01-01' or sale_date>'2019-03-31');

这篇关于力扣SQL仅数据库(1068~1084)的文章就介绍到这儿,希望我们推荐的文章对编程师们有所帮助!



http://www.chinasem.cn/article/1132639

相关文章

MySQL 中的 JSON 查询案例详解

《MySQL中的JSON查询案例详解》:本文主要介绍MySQL的JSON查询的相关知识,本文给大家介绍的非常详细,对大家的学习或工作具有一定的参考借鉴价值,需要的朋友参考下吧... 目录mysql 的 jsON 路径格式基本结构路径组件详解特殊语法元素实际示例简单路径复杂路径简写操作符注意MySQL 的 J

Windows 上如果忘记了 MySQL 密码 重置密码的两种方法

《Windows上如果忘记了MySQL密码重置密码的两种方法》:本文主要介绍Windows上如果忘记了MySQL密码重置密码的两种方法,本文通过两种方法结合实例代码给大家介绍的非常详细,感... 目录方法 1:以跳过权限验证模式启动 mysql 并重置密码方法 2:使用 my.ini 文件的临时配置在 Wi

MySQL重复数据处理的七种高效方法

《MySQL重复数据处理的七种高效方法》你是不是也曾遇到过这样的烦恼:明明系统测试时一切正常,上线后却频频出现重复数据,大批量导数据时,总有那么几条不听话的记录导致整个事务莫名回滚,今天,我就跟大家分... 目录1. 重复数据插入问题分析1.1 问题本质1.2 常见场景图2. 基础解决方案:使用异常捕获3.

SQL中redo log 刷⼊磁盘的常见方法

《SQL中redolog刷⼊磁盘的常见方法》本文主要介绍了SQL中redolog刷⼊磁盘的常见方法,将redolog刷入磁盘的方法确保了数据的持久性和一致性,下面就来具体介绍一下,感兴趣的可以了解... 目录Redo Log 刷入磁盘的方法Redo Log 刷入磁盘的过程代码示例(伪代码)在数据库系统中,r

mysql中的group by高级用法

《mysql中的groupby高级用法》MySQL中的GROUPBY是数据聚合分析的核心功能,主要用于将结果集按指定列分组,并结合聚合函数进行统计计算,下面给大家介绍mysql中的groupby用法... 目录一、基本语法与核心功能二、基础用法示例1. 单列分组统计2. 多列组合分组3. 与WHERE结合使

Mysql用户授权(GRANT)语法及示例解读

《Mysql用户授权(GRANT)语法及示例解读》:本文主要介绍Mysql用户授权(GRANT)语法及示例,具有很好的参考价值,希望对大家有所帮助,如有错误或未考虑完全的地方,望不吝赐教... 目录mysql用户授权(GRANT)语法授予用户权限语法GRANT语句中的<权限类型>的使用WITH GRANT

Mysql如何解决死锁问题

《Mysql如何解决死锁问题》:本文主要介绍Mysql如何解决死锁问题,具有很好的参考价值,希望对大家有所帮助,如有错误或未考虑完全的地方,望不吝赐教... 目录【一】mysql中锁分类和加锁情况【1】按锁的粒度分类全局锁表级锁行级锁【2】按锁的模式分类【二】加锁方式的影响因素【三】Mysql的死锁情况【1

SQL BETWEEN 的常见用法小结

《SQLBETWEEN的常见用法小结》BETWEEN操作符是SQL中非常有用的工具,它允许你快速选取某个范围内的值,本文给大家介绍SQLBETWEEN的常见用法,感兴趣的朋友一起看看吧... 在SQL中,BETWEEN是一个操作符,用于选取介于两个值之间的数据。它包含这两个边界值。BETWEEN操作符常用

MySQL索引的优化之LIKE模糊查询功能实现

《MySQL索引的优化之LIKE模糊查询功能实现》:本文主要介绍MySQL索引的优化之LIKE模糊查询功能实现,本文通过示例代码给大家介绍的非常详细,感兴趣的朋友一起看看吧... 目录一、前缀匹配优化二、后缀匹配优化三、中间匹配优化四、覆盖索引优化五、减少查询范围六、避免通配符开头七、使用外部搜索引擎八、分

MySql match against工具详细用法

《MySqlmatchagainst工具详细用法》在MySQL中,MATCH……AGAINST是全文索引(Full-Textindex)的查询语法,它允许你对文本进行高效的全文搜素,支持自然语言搜... 目录一、全文索引的基本概念二、创建全文索引三、自然语言搜索四、布尔搜索五、相关性排序六、全文索引的限制七