Question

1082. Sales Analysis I

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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

# 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

# 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;