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1158. Market Analysis I

中文文档

Description

Table: Users

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| user_id        | int     |
| join_date      | date    |
| favorite_brand | varchar |
+----------------+---------+
user_id is the primary key of this table.
This table has the info of the users of an online shopping website where users can sell and buy items.

 

Table: Orders

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| item_id       | int     |
| buyer_id      | int     |
| seller_id     | int     |
+---------------+---------+
order_id is the primary key of this table.
item_id is a foreign key to the Items table.
buyer_id and seller_id are foreign keys to the Users table.

 

Table: Items

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| item_id       | int     |
| item_brand    | varchar |
+---------------+---------+
item_id is the primary key of this table.

 

Write an SQL query to find for each user, the join date and the number of orders they made as a buyer in 2019.

Return the result table in any order.

The query result format is in the following example.

 

Example 1:

Input: 
Users table:
+---------+------------+----------------+
| user_id | join_date  | favorite_brand |
+---------+------------+----------------+
| 1       | 2018-01-01 | Lenovo         |
| 2       | 2018-02-09 | Samsung        |
| 3       | 2018-01-19 | LG             |
| 4       | 2018-05-21 | HP             |
+---------+------------+----------------+
Orders table:
+----------+------------+---------+----------+-----------+
| order_id | order_date | item_id | buyer_id | seller_id |
+----------+------------+---------+----------+-----------+
| 1        | 2019-08-01 | 4       | 1        | 2         |
| 2        | 2018-08-02 | 2       | 1        | 3         |
| 3        | 2019-08-03 | 3       | 2        | 3         |
| 4        | 2018-08-04 | 1       | 4        | 2         |
| 5        | 2018-08-04 | 1       | 3        | 4         |
| 6        | 2019-08-05 | 2       | 2        | 4         |
+----------+------------+---------+----------+-----------+
Items table:
+---------+------------+
| item_id | item_brand |
+---------+------------+
| 1       | Samsung    |
| 2       | Lenovo     |
| 3       | LG         |
| 4       | HP         |
+---------+------------+
Output: 
+-----------+------------+----------------+
| buyer_id  | join_date  | orders_in_2019 |
+-----------+------------+----------------+
| 1         | 2018-01-01 | 1              |
| 2         | 2018-02-09 | 2              |
| 3         | 2018-01-19 | 0              |
| 4         | 2018-05-21 | 0              |
+-----------+------------+----------------+

Solutions

SQL

SELECT user_id AS buyer_id,
    join_date,
    COUNT(order_id) AS orders_in_2019
FROM users AS u
    LEFT JOIN orders AS o ON u.user_id = o.buyer_id
    AND YEAR(order_date) = 2019
GROUP BY user_id;
SELECT
    user_id AS buyer_id,
    join_date,
    (
        SELECT
            COUNT(*)
        FROM
            orders AS o
        WHERE
            u.user_id = o.buyer_id AND YEAR(order_date) = 2019
    ) AS orders_in_2019
FROM
    users AS u;