Question

I’ve been using SQL for a few years, and this type of problem comes up here and there, and I haven’t found an answer. But perhaps I’ve been looking in the wrong places – I’m not really sure what to call it.

For the sake of brevity, let’s say I have a table with 3 columns: Customer, Order_Amount, Order_Date. Each customer may have multiple orders, with one row for each order with the amount and date.

My Question: Is there a simple way in SQL to get the DATE of the maximum order per customer?

I can get the amount of the maximum order for each customer (and which customer made it) by doing something like:

SELECT Customer, MAX(Order_Amount) FROM orders GROUP BY Customer;

But I also want to get the date of the max order, which I haven’t figured out a way to easily get. I would have thought that this would be a common type of question for a database, and would therefore be easy to do in SQL, but I haven’t found an easy way to do it yet. Once I add Order_Date to the list of columns to select, I need to add it to the Group By clause, which I don’t think will give me what I want.

Answer 1

Apart from self-join you can do:

SELECT o1.*
FROM orders o1 JOIN orders o2 ON o1.Customer = o2.Customer
GROUP BY o1.Customer, o1.Order_Amount
HAVING o1.Order_Amount = MAX(o2.Order_Amount);

There’s a good article reviewing various approaches.

And in Oracle, db2, Sybase, SQL Server 2005+ you would use RANK() OVER.

SELECT * FROM (
    SELECT *
    RANK() OVER (PARTITION BY Customer ORDER BY Order_Amount DESC) r
    FROM orders) o 
WHERE r = 1;

Note: If Customer has more than one order with maximum Order_Amount (i.e. ties), using RANK() function would get you all such orders; to get only first one, replace RANK() with ROW_NUMBER().