586

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               |

Explanation

The customer with number '3' has two orders, which is greater than either customer '1' or '2' because each of them  only has one order. 
So the result is customer_number '3'.

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

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Solution 1.

# OJ: https://leetcode.com/problems/customer-placing-the-largest-number-of-orders/
# Author: github.com/lzl124631x
SELECT
    customer_number
FROM
    orders
GROUP BY customer_number
ORDER BY COUNT(*) DESC
LIMIT 1