SQL - Remove duplicates to show the latest date record
sql, sql-server-2008-r2, t-sql
Solution
since `SQL Server 2008 r2` supports windows function,
SELECT Customerid,
purchasedate,
paymenttype,
delivery,
amount,
discountrate
FROM
(
SELECT Customerid,
purchasedate,
paymenttype,
delivery,
amount,
discountrate,
ROW_NUMBER() OVER (Partition By CustomerID
ORDER BY purchasedate DESC) rn
FROM Customer
) derivedTable
WHERE derivedTable.rn = 1
or by using `Common Table Expression`
WITH derivedTable
AS
(
SELECT Customerid,
purchasedate,
paymenttype,
delivery,
amount,
discountrate,
ROW_NUMBER() OVER (Partition By CustomerID
ORDER BY purchasedate DESC) rn
FROM Customer
)
SELECT Customerid,
purchasedate,
paymenttype,
delivery,
amount,
discountrate
FROM derivedTable
WHERE derivedTable.rn = 1
or by using join with subquery which works in other DBMS
SELECT a.*
FROM Customer a
INNER JOIN
(
SELECT CustomerID, MAX(purchasedate) maxDate
FROM Customer
GROUP BY CustomerID
) b ON a.CustomerID = b.CustomerID AND
a.purchasedate = b.maxDate
Problem
I have a view which ultimately I want to return 1 row per customer. Currently its a Select as follows; ``` SELECT Customerid, MAX(purchasedate) AS purchasedate, paymenttype, delivery, amount, discountrate FROM Customer GROUP BY Customerid, paymenttype, delivery, amount, discountrate ``` I was hoping the MAX(purchasedate) would work but when I do my groupings it breaks as sometimes there could be a discountrate, sometimes its NULL, paymenttype can differ for each customer also, is there anyway just to show the last purchase a customer makes?