SQL question about GROUP BY - sql

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().

Answer 2

There's no short-cut... the easiest way is probably to join to a sub-query:
SELECT
*
FROM
orders JOIN
(
SELECT Customer, MAX(Order_Amount) AS Max_Order_Amount
FROM orders
GROUP BY Customer
) maxOrder
ON maxOrder.Customer = orders.Customer
AND maxOrder.Max_Order_Amount = orders.Order_Amount

Answer 3

you will want to join on the same table...
SELECT Customer, order_date, amt
FROM orders o,
( SELECT Customer, MAX(Order_Amount) amt FROM orders GROUP BY Customer ) o2
WHERE o.customer = o2.customer
AND o.order_amount = o2.amt
;

Answer 4

Another approach for the collection:
WITH tempquery AS
(
SELECT
Customer
,Order_Amount
,Order_Date
,row_number() OVER (PARTITION BY Customer ORDER BY Order_Amount DESC) AS rn
FROM
orders
)
SELECT
Customer
,Order_Amount
,Order_Date
FROM
tempquery
WHERE
rn = 1

Answer 5

If your DB Supports CROSS APPLY you can do this as well, but it doesn't handle ties correctly
SELECT [....]
FROM Customer c
CROSS APPLY
(SELECT TOP 1 [...]
FROM Orders o
WHERE c.customerID = o.CustomerID
ORDER BY o.Order_Amount DESC) o
See this data.SE query

Answer 6

You could try something like this:
SELECT Customer, MAX(Order_Amount), Order_Date
FROM orders O
WHERE ORDER_AMOUNT = (SELECT MAX(ORDER_AMOUNT) FROM orders WHERE CUSTOMER = O.CUSTOMER)
GROUP BY CUSTOMER, Order_Date

Answer 7

with t as
(
select CUSTOMER,Order_Date ,Order_Amount,max(Order_Amount) over (partition
by Customer) as
max_amount from orders
)
select * from t where t.Order_Amount=max_amount