Sum of orders per customer ID

So, I have two tables, customers and orders and I want to get all orders associated with a customer.

The tables look like this:

Customers

customer_no order_no
1027 V327
1027 V391
1028 V400
etc..


Orders

order_no article1 article2 size order_date quantity
V327 Jeans Black 36 20150101 2
V327 Shirt Blue L 20150101 1
V327 Shirt White L 20150101 2
V391 Jeans Red 34 20150202 3
V400 Shirt Green 32 20150226 2
etc..


Now, I currently use the following query, but it is dog slow and I just know I'm approaching this wrongly. I feel like it should be possible to solve this with a join somewhere, but for the life of mine, I just cannot figure this one out.

SELECT article1, article2, size, order_date, SUM(quantity)
FROM orders
WHERE order_no IN (
SELECT order_no
FROM customers
WHERE customer_no = '1027'
)
GROUP BY article1, article2, size, order_date, SUM(quantity)


The output for customer 1027 is (and should be):

article1 article2 size order_date SUM(quantity)
Jeans Black 36 20150101 2
Shirt Blue L 20150101 1
Shirt White L 20150101 2
Jeans Red 34 20150202 3


Does anyone have a better way of solving this? Performance is my main issue here.

Applying a JOIN sounds pretty easy

SELECT article1, article2, size, order_date, SUM(quantity)
FROM orders 
INNER JOIN customers ON orders.order_no = customers.order_no
WHERE customers.customer_no = '1027'
GROUP BY article1, article2, size, order_date, SUM(quantity)

You should note that nested SELECT statements in a WHERE clause almost always lead to bad performance.

Also that kind of unnormalized data model looks weird, but there may be good reasons why to keep order_no and customer_no pairs in a single association table. May be its just bad naming, something like customers_orders might probably have been a better choice.

StackExchange Code Review Q#114750, answer score: 5