# Sellers With No Sales Problem

## Description

LeetCode Problem 1607.

Table: Customer

``````1
2
3
4
5
6
7
8
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| customer_name | varchar |
+---------------+---------+
customer_id is the primary key for this table.
Each row of this table contains the information of each customer in the WebStore.
``````

Table: Orders

``````1
2
3
4
5
6
7
8
9
10
11
12
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| sale_date     | date    |
| order_cost    | int     |
| customer_id   | int     |
| seller_id     | int     |
+---------------+---------+
order_id is the primary key for this table.
Each row of this table contains all orders made in the webstore.
sale_date is the date when the transaction was made between the customer (customer_id) and the seller (seller_id).
``````

Table: Seller

``````1
2
3
4
5
6
7
8
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| seller_id     | int     |
| seller_name   | varchar |
+---------------+---------+
seller_id is the primary key for this table.
Each row of this table contains the information of each seller.
``````

Write an SQL query to report the names of all sellers who did not make any sales in 2020.

Return the result table ordered by seller_name in ascending order.

The query result format is in the following example.

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
Customer table:
+--------------+---------------+
| customer_id  | customer_name |
+--------------+---------------+
| 101          | Alice         |
| 102          | Bob           |
| 103          | Charlie       |
+--------------+---------------+

Orders table:
+-------------+------------+--------------+-------------+-------------+
| order_id    | sale_date  | order_cost   | customer_id | seller_id   |
+-------------+------------+--------------+-------------+-------------+
| 1           | 2020-03-01 | 1500         | 101         | 1           |
| 2           | 2020-05-25 | 2400         | 102         | 2           |
| 3           | 2019-05-25 | 800          | 101         | 3           |
| 4           | 2020-09-13 | 1000         | 103         | 2           |
| 5           | 2019-02-11 | 700          | 101         | 2           |
+-------------+------------+--------------+-------------+-------------+

Seller table:
+-------------+-------------+
| seller_id   | seller_name |
+-------------+-------------+
| 1           | Daniel      |
| 2           | Elizabeth   |
| 3           | Frank       |
+-------------+-------------+

Result table:
+-------------+
| seller_name |
+-------------+
| Frank       |
+-------------+
Daniel made 1 sale in March 2020.
Elizabeth made 2 sales in 2020 and 1 sale in 2019.
Frank made 1 sale in 2019 but no sales in 2020.
``````

## MySQL Solution

``````1
2
3
4
5
6
7
8
9
10
select sellInfo.seller_name as seller_name
from (
select s.seller_name as seller_name, max(o.sale_date) as latest_sale_date
from Orders o
right join Seller s
on o.seller_id = s.seller_id
group by s.seller_id
) sellInfo
where sellInfo.latest_sale_date is null or datediff('2020-01-01', sellInfo.latest_sale_date) > 0
order by seller_name
``````