1082. Sales Analysis I 🔒

Description

Table: Product

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| product_id   | int     |
| product_name | varchar |
| unit_price   | int     |
+--------------+---------+
product_id is the primary key (column with unique values) 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 can have repeated rows.
product_id is a foreign key (reference column) to the Product table.
Each row of this table contains some information about one sale.

 

Write a solution that reports the best seller by total sales price, If there is a tie, report them all.

Return the result table in any order.

The 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: 
+-------------+
| seller_id   |
+-------------+
| 1           |
| 3           |
+-------------+
Explanation: Both sellers with id 1 and 3 sold products with the most total price of 2800.

Solutions

Solution 1

MySQL

# Write your MySQL query statement below
SELECT seller_id
FROM Sales
GROUP BY seller_id
HAVING
    SUM(price) >= ALL(
        SELECT SUM(price)
        FROM Sales
        GROUP BY seller_id
    );

Solution 2

MySQL

# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            seller_id,
            SUM(price) AS tot,
            RANK() OVER (ORDER BY SUM(price) DESC) AS rk
        FROM Sales
        GROUP BY seller_id
    )
SELECT seller_id
FROM T
WHERE rk = 1;

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