I am learning SQL server and I am stuck on a question.
I need to write a query that shows each customers last order that he placed and the order before the last one he made.
Thank you for your help!
Edit: So far I have this:
SELECT SalesOrderID, CustomerID, per.FirstName, per.LastName, OrderDate as "Latest Order Date"
FROM (
SELECT *,
Row_Number() OVER (PARTITION BY CustomerID ORDER BY OrderDate desc) as 'Rank'
FROM sales.SalesOrderHeader head
) a join Person.Person per
on CustomerID = per.BusinessEntityID
WHERE Rank = 1
As you can see, I am pretty close. I just need to add a column that shows the order before the latest order date.
Sorry, I'm new to the site (long time viewer, first time poster)
ty!
You need to combine both ROW_NUMBER and LEAD
LEAD is better in this case, because LAG needs the rows sorted in the opposite direction from the ROW_NUMBER
SELECT head.SalesOrderID, CustomerID, per.FirstName, per.LastName, OrderDate as LastOrder, head.PreviousOrder
FROM (
SELECT *,
ROW_NUMBER() OVER (PARTITION BY CustomerID ORDER BY OrderDate DESC) as rnk,
LEAD(OrderDate) OVER (PARTITION BY CustomerID ORDER BY OrderDate DESC) as PreviousOrder
FROM sales.SalesOrderHeader head
) head
JOIN Person.Person per ON head.CustomerID = per.BusinessEntityID
WHERE head.rnk = 1;
Related
select * from
(select Id, Prodcut, Billing_date
, row_number() over (partition by Id, product order by Billing_date desc) as RowNumber
,sum(Revenue)
from Table1
group by 1,2,3,4,1) a
where a.rowNumber = 1
There are rows where Id+product combination repeats for latest billing date and which causing some data to be missed out. I am trying to add sum with row_number to sum all the ID&product combinations for the latest date but not able to make it work.
Can anyone please help me out here!
Data Sample Image
Database: Athena, Dbeaver
I would expect this to do what you want:
select *
from (select Id, Product, Billing_date,
row_number() over (partition by Id, product order by Billing_date desc) as seqnum,
sum(Revenue)
from Table1
group by Id, Product, Billing_date
) t1
where seqnum = 1;
Your group by columns do not seem correct. I'm surprised your query runs in any datbase.
I have the following query to pull a customer's most recent purchase. I tried to use a subselect for performance reasons, but I ran into a wall and kept getting back ALL the customers' orders. I just need the most recent for each individual customer.
SELECT *
FROM (SELECT od.*, ord.OrderName, ord.OrderDate, RowN =
Row_Number()
OVER (PARTITION BY ord.CustomerOrderGUID ORDER BY ord.OrderDate DESC)
FROM #OrderData od
JOIN CV3Orders ord ON ord.CustomerOrderGUID = od.CustomerOrderGUID
WHERE ord.ProductName = 'Product 10') rnk
WHERE rnk.RowN = 1
CustomerOrderGuid would seem to represent every order not every customer. So, you need to partition by the correct column. I might guess:
SELECT co.*
FROM (SELECT od.*, ord.OrderName, ord.OrderDate,
Row_Number() OVER (PARTITION BY ord.CustomerGUID ORDER BY ord.OrderDate DESC) as seqnum
FROM #OrderData od JOIN
CV3Orders ord
ON ord.CustomerOrderGUID = od.CustomerOrderGUID
WHERE ord.ProductName = 'Product 10'
) co
WHERE seqnum = 1;
It's an orcale database and I'm trying to get the last order ID and its date for each customer. I'm familiar enough with MySQL to be able to write something like the following in a MySQL system:
https://www.w3schools.com/SQL/trysql.asp?filename=trysql_select_all
SELECT CustomerID, COUNT(OrderID), MAX(OrderDate) as Last_Order_Date,
(SELECT OrderID FROM Orders O2
WHERE O2.CustomerId = O1.CustomerId
ORDER BY OrderDate DESC LIMIT 1)
AS LAST_ORDER_ID
FROM [Orders] O1
GROUP BY CustomerID
ORDER BY CustomerID
However I haven't been able to write the equivalent for an Oracle database. The part that trips me up is I have to write a subquery for the order by, then pick the top 1... but that also need to go into my bigger query of getting the last order for each customer.
Use MAX( ... ) KEEP ( DENSE_RANK LAST ... ):
SELECT CustomerID,
COUNT( OrderID ),
MAX( OrderDate ) AS Last_Order_Date,
MAX( OrderID ) KEEP ( DENSE_RANK LAST ORDER BY OrderDate ) AS Last_Order_ID
FROM Orders
GROUP BY CustomerID
ORDER BY CustomerID;
or use analytic functions:
SELECT CustomerID,
Num_Orders,
OrderDate AS Last_Order_Date,
OrderID AS Last_Order_ID
FROM (
SELECT o.*,
COUNT( OrderID ) OVER ( PARTITION BY CustomerID ) AS Num_Orders,
ROW_NUMBER() OVER ( PARTITION BY CustomerID
ORDER BY OrderDate DESC, OrderID DESC ) AS rn
FROM orders o
)
WHERE rn = 1
ORDER BY CustomerID;
I have three sets of information:
productID
date
seller
I need to get information of last product sold for each productID and sold by. I tried using max value of date but that forces me to use grouping for seller as well but I don't want to group by seller. I want to group by productID, get the date it was sold last and by who. How can I avoid grouping on seller?
Use Window function which will help you to find the Latest date in each group(productId)
SELECT ProductID,
[date],
seller
FROM (SELECT Row_number()
OVER(
partition BY ProductID
ORDER BY [date] desc) Rn,
*
FROM tablename) a
WHERE rn = 1
or use can also use Max aggregate with group by to get the result
SELECT ProductID,
[date],
seller
FROM tablename a
JOIN (SELECT Max([date]) [date],
productid
FROM tablename
group by productid) b
ON a.productid = b.productid
AND a.[date] = b.[date]
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.
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().
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
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
;
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
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
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
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