1068. 产品销售分析1
需求
sql
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
sql
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
sql
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
sql
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
sql
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
sql
select product_id,sum(quantity) total_quantity from sales group by product_id;1070. 产品销售分析3
需求
sql
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
sql
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
sql
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
sql
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
sql
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
sql
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
sql
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
sql
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)
sql
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
sql
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
sql
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
sql
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
sql
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
sql
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
sql
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
sql
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
sql
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
sql
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' and
buyer_id not in (select buyer_id from t1 where product_name ='iPhone');1084. 销售分析3
需求
sql
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
sql
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
sql
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');