Question

Consider following schema:

 Customers: Col      | Type    | -------------------| id       | INTEGER | name     | VARCHAR |  Orders: Col         | Type    | ----------------------| id          | INTEGER | customer_id | INTEGER | date        | DATE    |  Items Col         | Type    | ----------------------| id          | INTEGER | order_id    | INTEGER | name        | VARCHAR | quantity    | INTEGER | 

And here’s the breif run down of the schema: Customer can have many orders, and an order can have many items Note: You may argue that the items should refer to some products table with a product_id, but let it be simple as is now.

From my database, I want to ask following question:

For a given customer, list all the items that start with La.

My version of the query goes like this:

SELECT * FROM items WHERE name LIKE ‘La%’

This would return ALL the items that Begin with La for ALL the customers. What I want is that I get all the items for a SPECIFIC customer with say id 1. How do I write such a query? My second version that would do is like this:

SELECT * FROM items WHERE name LIKE ‘La%’ AND order_id in (SELECT id FROM orders WHERE customer_id=1);

But is their any other efficient/elegant or a more better way to do the same thing?

Answer 1

The elegant way to do this is called a join.

SELECT   c.id            customer_id,   c.name          customer_name,   i.id            item_id,   i.name          item_name,   COUNT(i.id)     order_count,   SUM(i.quantity) items_ordered FROM   customers c   INNER JOIN orders o ON c.id = o.customer_id   INNER JOIN items  i ON i.id = o.order_id WHERE   i.name LIKE 'La%'   AND c.id = 1 GROUP BY   c.id,   c.name,   i.id,   i.name