Market Analysis I Problem
Description
LeetCode Problem 1158.
Table: Users
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+----------------+---------+
| 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
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+---------------+---------+
| 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
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+---------------+---------+
| 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.
The query result format is in the following example:
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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 |
+---------+------------+
Result table:
+-----------+------------+----------------+
| 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 |
+-----------+------------+----------------+
MySQL Solution
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select user_id as buyer_id, join_date,
sum(case when order_id is not null then 1 else 0 end) as orders_in_2019
from users
left join orders on users.user_id = orders.buyer_id
and year(order_date) = 2019
group by user_id, join_date
order by buyer_id
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