# 1532. The Most Recent Three Orders

<https://leetcode.com/problems/the-most-recent-three-orders>

## Description

Table: `Customers`

```
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer\_id   | int     |
| name          | varchar |
+---------------+---------+
customer\_id is the primary key for this table.
This table contains information about customers.
```

Table: `Orders`

```
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order\_id      | int     |
| order\_date    | date    |
| customer\_id   | int     |
| cost          | int     |
+---------------+---------+
order\_id is the primary key for this table.
This table contains information about the orders made by customer\_id.
Each customer has **one order per day**.
```

Write an SQL query to find the most recent 3 orders of each user. If a user ordered less than 3 orders return all of their orders.

Return the result table sorted by `customer_name` in **ascending** order and in case of a tie by the `customer_id` in **ascending** order. If there still a tie, order them by the `order_date` in **descending** order.

The query result format is in the following example:

```
Customers
+-------------+-----------+
| customer\_id | name      |
+-------------+-----------+
| 1           | Winston   |
| 2           | Jonathan  |
| 3           | Annabelle |
| 4           | Marwan    |
| 5           | Khaled    |
+-------------+-----------+
Orders
+----------+------------+-------------+------+
| order\_id | order\_date | customer\_id | cost |
+----------+------------+-------------+------+
| 1        | 2020-07-31 | 1           | 30   |
| 2        | 2020-07-30 | 2           | 40   |
| 3        | 2020-07-31 | 3           | 70   |
| 4        | 2020-07-29 | 4           | 100  |
| 5        | 2020-06-10 | 1           | 1010 |
| 6        | 2020-08-01 | 2           | 102  |
| 7        | 2020-08-01 | 3           | 111  |
| 8        | 2020-08-03 | 1           | 99   |
| 9        | 2020-08-07 | 2           | 32   |
| 10       | 2020-07-15 | 1           | 2    |
+----------+------------+-------------+------+
Result table:
+---------------+-------------+----------+------------+
| customer\_name | customer\_id | order\_id | order\_date |
+---------------+-------------+----------+------------+
| Annabelle     | 3           | 7        | 2020-08-01 |
| Annabelle     | 3           | 3        | 2020-07-31 |
| Jonathan      | 2           | 9        | 2020-08-07 |
| Jonathan      | 2           | 6        | 2020-08-01 |
| Jonathan      | 2           | 2        | 2020-07-30 |
| Marwan        | 4           | 4        | 2020-07-29 |
| Winston       | 1           | 8        | 2020-08-03 |
| Winston       | 1           | 1        | 2020-07-31 |
| Winston       | 1           | 10       | 2020-07-15 |
+---------------+-------------+----------+------------+
Winston has 4 orders, we discard the order of "2020-06-10" because it is the oldest order.
Annabelle has only 2 orders, we return them.
Jonathan has exactly 3 orders.
Marwan ordered only one time.
We sort the result table by customer\_name in ascending order, by customer\_id in ascending order and by order\_date in descending order in case of a tie.
```

**Follow-up:**\
Can you write a general solution for the most recent `n` orders?

## ac

```java
```


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