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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
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# 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 );