Get top one record of difference date - sql

Question

I try to get just one record of last Order of everysingle customer. ATM I have this idea but it giving me all of them Date of orders.
I work on dataBase AdventurerWorks2012
SELECT
H.CustomerID AS Customer,
H.SalesOrderID AS OrderNumber,
MAX(CAST(H.OrderDate AS DATE)) AS DateOrder
FROM
Sales.SalesOrderHeader H
JOIN
Sales.SalesOrderDetail D
ON H.SalesOrderID = D.SalesOrderID
GROUP BY
H.CustomerID,
H.SalesOrderID
ORDER BY
CustomerID;

Answer 1

You can use TOP 1 WITH TIES
select top 1 with ties
*
From
Sales.SalesOrderHeader H
INNER JOIN Sales.SalesOrderDetail D
ON H.SalesOrderID = D.SalesOrderID
ORDER BY row_number() over (partition by H.CustomerID order by H.OrderDate desc)
This applies and ordering with row_number() giving each customer's order an ID starting at 1 and going to N based off the OrderDate. The WITH TIES allows us to return the TOP 1 for each customer.
Another way, is using a CTE
;with cte as(
select top 1 with ties
*, RN = row_number() over (partition by H.CustomerID order by H.OrderDate desc)
From
Sales.SalesOrderHeader H
INNER JOIN Sales.SalesOrderDetail D
ON H.SalesOrderID = D.SalesOrderID)
select *
from cte
where RN = 1

Answer 2

I think you don't need to do JOIN:
SELECT h.*
FROM Sales.SalesOrderHeader h
WHERE OrderDate = (SELECT MAX(h1.OrderDate)
FROM Sales.SalesOrderHeader h1
WHERE h.CustomerID = h1.CustomerID
);
By this you will get customers with most recent order information.

Answer 3

select top 1 with ties
H.CustomerID AS Customer,
H.SalesOrderID as OrderNumber,
H.OrderDate As DataLast
From
Sales.SalesOrderHeader H
INNER JOIN Sales.SalesOrderDetail D
ON H.SalesOrderID = D.SalesOrderID
ORDER BY row_number() over (partition by H.CustomerID order by H.OrderDate desc)
This work like i need. Thanks!