Algorithm

Problem Name: 1084. Sales Analysis III

Problem Link: https://leetcode.com/problems/sales-analysis-iii/
 
SQL Schema

Table: Product

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| product_id   | int     |
| product_name | varchar |
| unit_price   | int     |
+--------------+---------+
product_id is the primary key of this table.
Each row of this table indicates the name and the price of each product.

Table: Sales

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| seller_id   | int     |
| product_id  | int     |
| buyer_id    | int     |
| sale_date   | date    |
| quantity    | int     |
| price       | int     |
+-------------+---------+
This table has no primary key, it can have repeated rows.
product_id is a foreign key to the Product table.
Each row of this table contains some information about one sale.

 

Write an SQL query that reports the products that were only sold in the first quarter of 2019. That is, between 2019-01-01 and 2019-03-31 inclusive.

Return the result table in any order.

The query result format is in the following example.

 

Example 1:

Input: 
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  |
+-----------+------------+----------+------------+----------+-------+
Output: 
+-------------+--------------+
| product_id  | product_name |
+-------------+--------------+
| 1           | S8           |
+-------------+--------------+
Explanation: 
The product with id 1 was only sold in the spring of 2019.
The product with id 2 was sold in the spring of 2019 but was also sold after the spring of 2019.
The product with id 3 was sold after spring 2019.
We return only product 1 as it is the product that was only sold in the spring of 2019.

Code Examples

#1 Code Example with SQL

Code - SQL


# Write your MySQL query statement below
SELECT product_id, product_name
FROM Sales 
JOIN Product 
Using(product_id)
GROUP BY product_id
HAVING MIN(sale_date) >= '2019-01-01' AND MAX(sale_date) <= '2019-03-31'
Copy The Code & Try With Live Editor

Input

x
–
+
cmd
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 | +-----------+------------+----------+------------+----------+-------+

Output

x
–
+
cmd
+-------------+--------------+ | product_id | product_name | +-------------+--------------+ | 1 | S8 | +-------------+--------------+
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Demonstration

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