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力扣SQL仅数据库（1068~1084）

news 来源：原创 2024/9/20 9:33:32

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：

使用子查询得出最大值
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');