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?
Solution 1:
# Write your MySQL query statement below
SELECT customer_number
FROM
(SELECT customer_number, count(customer_number) as num
FROM orders
GROUP BY customer_number) AS tmp
ORDER BY num DESC
LIMIT 1
Solution 2:
# Write your MySQL query statement below
SELECT customer_number
FROM
(SELECT customer_number, count(customer_number) as num
FROM orders
GROUP BY customer_number
ORDER BY num DESC
LIMIT 1
) AS tmp
Solution 3:
# Write your MySQL query statement below
SELECT customer_number from orders
GROUP BY customer_number
ORDER BY count(customer_number) DESC
LIMIT 1
Summary:
- Nothing to say.
LeetCode: 586. Customer Placing the Largest Number of Orders
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