586_Customer Placing the Largest Number of Orders

[easy]

Query the customer_number from the orders table for the customer who has placed the largest number of orders. It is guaranteed that exactly one customer will have placed more orders than any other customer. The orders table is defined as follows:

| Column            | Type      |
|-------------------|-----------|
| order_number (PK) | int       |
| customer_number   | int       |
| order_date        | date      |
| required_date     | date      |
| shipped_date      | date      |
| status            | char(15)  |
| comment           | char(200) |

Sample Input

| order_number | customer_number | order_date | required_date | shipped_date | status | comment |
|--------------|-----------------|------------|---------------|--------------|--------|---------|
| 1            | 1               | 2017-04-09 | 2017-04-13    | 2017-04-12   | Closed |         |
| 2            | 2               | 2017-04-15 | 2017-04-20    | 2017-04-18   | Closed |         |
| 3            | 3               | 2017-04-16 | 2017-04-25    | 2017-04-20   | Closed |         |
| 4            | 3               | 2017-04-18 | 2017-04-28    | 2017-04-25   | Closed |         |

Sample Output

| customer_number |
|-----------------|
| 3               |

Solution:

SELECT customer_number
FROM orders
GROUP BY customer_number
ORDER BY COUNT(*) DESC
LIMIT 1;

Follow up:

What if more than one customer have the largest number of orders, can you find all the customer_number in this case?

SELECT customer_number
FROM orders
GROUP BY customer_number
HAVING COUNT(order_number) = 
    (
    SELECT COUNT(order_number)
    FROM orders
    GROUP BY customer_number
    ORDER BY COUNT(order_number) DESC
    LIMIT 1
    );