Question

I have a customers table, and an orders table. Each customer can have many orders.

I want to select every customer, along with their earliest order number from the orders table (it is important that I select the earliest order number, not just any order). I want to list customers whether they have an order or not, and I don’t want to include customers twice if they have more than one order.

I’m using this:

  SELECT *
    FROM customers
    LEFT JOIN orders
    ON customers.id = orders.customer_id 
    GROUP BY customers.id

This gives me almost what I want, except it will pick whatever order ID it likes from the table. I need to be able to sort, and pick the smallest order ID.

Any ideas please?

Im pretty sure its something that’s staring me in the face…

EDIT: Tables’ Structures as requested

• Customers:

  ---

  | ID | Name | Address            | Etc |
  ----------------------------------------
  | 1  | Joe  | 123 Fake Street    |     |
  | 2  | Mike | 1600 Fake Road     |     |
  | 3  | Bill | Red Square, Moscow |     |
  ----------------------------------------
  

• Orders:

  ---

  | ID | Customer_ID | Date |
  ---------------------------
  | 1  | 1           |  ... |
  | 2  | 2           |  ... |
  | 3  | 2           |  ... |
  | 4  | 1           |  ... |
  ---------------------------
  

Answer 1

Create a virtual table (a/k/a subquery) with the lowest numerical order ID for each customer.

SELECT customer_id, min(order_id)
  FROM orders
 GROUP BY customer_id

Then join that table with the customer table, like so.

SELECT C.customer_id, firstorder.order_id
  FROM CUSTOMERS as C
  LEFT JOIN (
    SELECT customer_id, min(order_id)
      FROM orders
     GROUP BY customer_id
  ) AS firstorder ON c.customer_id = firstorder.customer_id