SQL: improving join efficiency - sql

If I turn this sub-query which selects sales persons and their highest price paid for any item they sell:
select *,
(select top 1 highestProductPrice
from orders o
where o.salespersonid = s.id
order by highestProductPrice desc ) as highestProductPrice
from salespersons s
in to this join in order to improve efficiency:
select *, highestProductPrice
from salespersons s join (
select salespersonid, highestProductPrice, row_number(
partition by salespersonid
order by salespersonid, highestProductPrice) as rank
from orders ) o on s.id = o.salespersonid
It still touches every order record (it enumerates the entire table before filtering by salespersonid it seems.) However you cannot do this:
select *, highestProductPrice
from salespersons s join (
select salespersonid, highestProductPrice, row_number(
partition by salespersonid
order by salespersonid, highestProductPrice) as rank
from orders
where orders.salepersonid = s.id) o on s.id = o.salespersonid
The where clause in the join causes a `multi-part identifier "s.id" could not be bound.
Is there any way to join the top 1 out of each order group with a join but without touching each record in orders?

Try
SELECT
S.*,
T.HighestProductPrice
FROM
SalesPersons S
CROSS APPLY
(
SELECT TOP 1 O.HighestProductPrice
FROM Orders O
WHERE O.SalesPersonid = S.Id
ORDER BY O.SalesPersonid, O.HighestProductPrice DESC
) T

would
select s.*, max(highestProductPrice)
from salespersons s
join orders o on o.salespersonid = s.id
group by s.*
or
select s.*, highestProductPrice
from salespersons s join (select salepersonid,
max(highestProductPrice) as highestProductPrice
from orders o) as o on o.salespersonid = s.id
work?

Related

Display Product Name and City where that product sold in largest quantity

I'm trying to get a query to display the product name and city where the product had the highest quantity sold. Here is the code I'm working with:
SELECT DISTINCT
(s.city),
MAX(t.quantity),
p.Name
FROM [DS715-Cameron-Erwin].dbo.Tb_Transactions AS t,
[DS715-Cameron-Erwin].dbo.Tb_Product AS p,
[DS715-Cameron-Erwin].dbo.Tb_Supplier AS s
WHERE p.prod_id = t.prod_id
AND s.Supp_ID = t.Supp_ID
GROUP BY t.Prod_ID,
p.name,
s.city
ORDER BY p.name, s.city
This is giving me the highest quantity sold for each product in each city.
Sample Data
From the screenshot there are multiple records for each product (Airplane, Auto, Boat...). I'm trying to get a single record for each product where ever the highest quantity was purchased.
So, the top record would only show for Airplane because the most orders were from there.
You want to use the ROW_NUMBER() OVER functionality to order by the quantity and then select the one with the biggest quantity over each product.
SELECT
city,
quantity,
name
FROM
(
SELECT S.city,
T.quantity,
P.name,
ROW_NUMBER() OVER
( PARTITION BY
P.name
ORDER BY t.Quantity DESC
) as RowNum
FROM
Tb_Transactions T
INNER JOIN
Tb_Product P
ON
P.prod_id = T.prod_id
INNER JOIN
Tb_Supplier S
ON
S.supp_id = T.supp_id
) a
WHERE
RowNum = 1
http://sqlfiddle.com/#!6/628458/5
For this, I would use a CTE (also I would use the explicit INNER JOIN syntax):
;With CTE
As
(
Select
s.city
, t.quantity
, p.Name
, Row_Number Over (Partition By P.Name, s.city Order By t.Quantity Desc) as RN
From [DS715-Cameron-Erwin].dbo.Tb_Transactions as t
Inner Join [DS715-Cameron-Erwin].dbo.Tb_Product as p
On p.prod_id = t.prod_id
Inner Join [DS715-Cameron-Erwin].dbo.Tb_Supplier as s
On s.Supp_ID = t.Supp_ID
)
Select
city
, quantity
, Name
From CTE
Where RN = 1

SQL Query to return the Top 2 Values

I am trying to return the top 2 most ordered items in our customer database. Below is what I have for the most ordered item but I am having trouble figuring out how to create another column for the 2nd most ordered item.
What is the best way to create the 2nd column?
SELECT FirstName, EmailAddress, Id, PreferredLocationId,
(
SELECT TOP 1 [Description] FROM [Order] o
INNER JOIN [OrderItem] oi ON oi.OrderId = o.OrderId
WHERE o.CustomerId = Customer.Id
GROUP BY [Description]
ORDER BY COUNT(*) DESC
) AS MostOrderedItem
FROM Customer
GROUP BY FirstName, EmailAddress, Id, PreferredLocationId
Lot's of different ways to handle this if you're using SQL Server 2012. I'm going to use a CTE to get the first two rows and use ROW_NUMBER()
WITH cte AS (
SELECT CustomerId, [Description]
, ROW_NUMBER() OVER (PARTITION BY CustomerId ORDER BY COUNT(*) DESC) [RowID]
FROM [Order] o
INNER JOIN [OrderItem] oi ON oi.OrderId = o.OrderId
GROUP BY CustomerId, [Description]
)
SELECT FirstName, EmailAddress, Id, PreferredLocationId, cte1.Description, cte2.Description
FROM Customer
LEFT JOIN cte cte1 ON cte1.CustomerID = Customer.CustomerId AND cte1.RowID = 1
LEFT JOIN cte cte2 ON cte2.CustomerID = Customer.CustomerId AND cte2.RowID = 2
The Common Table Expression creates the list of all customers, descriptions and their row number. Note that if you have ties, you're not guarunteed which description will come first. You can add to to the windowing function description so that if there is a tie, whatever comes first in the alphabet will be the tie breaker.

Last order item in Oracle SQL

I need to list columns from customer table, the date from first order and all data from last one, in a 1:N relationship between customer and order tables. I'm using Oracle 10g.
How the best way to do that?
TABLE CUSTOMER
---------------
id NUMBER
name VARCHAR2(200)
subscribe_date DATE
TABLE ORDER
---------------
id NUMBER
id_order NUMBER
purchase_date DATE
purchase_value NUMBER
Here is one way of doing it, using the row_number function, one join, and on aggregation:
select c.*,
min(o.purchase_date) as FirstPurchaseDate,
min(case when seqnum = 1 then o.id_order end) as Last_IdOrder,
min(case when seqnum = 1 then o.purchase_date end) as Last_PurchaseDate,
min(case when seqnum = 1 then o.purchase_value end) as Last_PurchaseValue
from Customer c join
(select o.*,
row_number() over (partition by o.id order by purchase_date desc) as seqnum
from orders o
) o
on c.customer_id = o.order_id
group by c.customer_id, c.name, c.subscribe_date
It's not obvious how to join the customer table to the orders table (order is a reserved word in Oracle so your table can't be named order). If we assume that the id_order in orders joins to the id in customer
SELECT c.id customer_id,
c.name name,
c.subscribe_date,
o.first_purchase_date,
o.id last_order_id,
o.purchase_date last_order_purchase_date,
o.purchase_value last_order_purchase_value
FROM customer c
JOIN (SELECT o.*,
min(o.purchase_date) over (partition by id_order) first_purchase_date,
rank() over (partition by id_order order by purchase_date desc) rnk
FROM orders o) o ON (c.id = o.id_order)
WHERE rnk = 1
I'm confused by your field names, but I'm going to assume that ORDER.id is the id in the CUSTOMER table.
The earliest order date is easy.
select CUSTOMER.*, min(ORDER.purchase_date)
from CUSTOMER
inner join ORDER on CUSTOMER.id = ORDER.id
group by CUSTOMER.*
To get the last order data, join this to the ORDER table again.
select CUSTOMER.*, min(ORD_FIRST.purchase_date), ORD_LAST.*
from CUSTOMER
inner join ORDER ORD_FIRST on CUSTOMER.id = ORD_FIRST.id
inner join ORDER ORD_LAST on CUSTOMER.id = ORD_LAST.id
group by CUSTOMER.*, ORD_LAST.*
having ORD_LAST.purchase_date = max(ORD_FIRST.purchase_date)
Maybe something like this assuming the ID field in the Order table is actually the Customer ID:
SELECT C.*, O1.*, O2.purchase_Date as FirstPurchaseDate
FROM Customer C
LEFT JOIN
(
SELECT Max(purchase_date) as pdate, id
FROM Orders
GROUP BY id
) MaxPurchaseOrder
ON C.Id = MaxPurchaseOrder.Id
LEFT JOIN Orders O1
ON MaxPurchaseOrder.pdate = O1.purchase_date
AND MaxPurchaseOrder.id = O1.id
LEFT JOIN
(
SELECT Min(purchase_date) as pdate, id
FROM Orders
GROUP BY id
) MinPurchaseOrder
ON C.Id = MinPurchaseOrder.Id
LEFT JOIN Orders O2
ON MinPurchaseOrder.pdate = O2.purchase_date
AND MinPurchaseOrder.id = O2.id
And the sql fiddle.

SQL Query for counting number of orders per customer and Total Dollar amount

I have two tables
Order with columns:
OrderID,OrderDate,CID,EmployeeID
And OrderItem with columns:
OrderID,ItemID,Quantity,SalePrice
I need to return the CustomerID(CID), number of orders per customer, and each customers total amount for all orders.
So far I have two separate queries. One gives me the count of customer orders....
SELECT CID, Count(Order.OrderID) AS TotalOrders
FROM [Order]
Where CID = CID
GROUP BY CID
Order BY Count(Order.OrderID) DESC;
And the other gives me the total sales. I'm having trouble combining them...
SELECT CID, Sum(OrderItem.Quantity*OrderItem.SalePrice) AS TotalDollarAmount
FROM OrderItem, [Order]
WHERE OrderItem.OrderID = [Order].OrderID
GROUP BY CID
I'm doing this in Access 2010.
You would use COUNT(DISTINCT ...) in other SQL engines:
SELECT CID,
Count(DISTINCT O.OrderID) AS TotalOrders,
Sum(OI.Quantity*OI.SalePrice) AS TotalDollarAmount
FROM [Order] O
INNER JOIN [OrderItem] OI
ON O.OrderID = OI.OrderID
GROUP BY CID
Order BY Count(DISTINCT O.OrderID) DESC
Which Access unfortunately does not support. Instead you can first get the Order dollar amounts and then join them before figuring the order counts:
SELECT CID,
COUNT(Orders.OrderID) AS TotalOrders,
SUM(OrderAmounts.DollarAmount) AS TotalDollarAmount
FROM [Orders]
INNER JOIN (SELECT OrderID, Sum(Quantity*SalePrice) AS DollarAmount
FROM OrderItems GROUP BY OrderID) AS OrderAmounts
ON Orders.OrderID = OrderAmounts.OrderID
GROUP BY CID
ORDER BY Count(Orders.OrderID) DESC
If you need to include Customers that have orders with no items (unusual but possible), change INNER JOIN to LEFT OUTER JOIN.
Create a query which uses your 2 existing queries as subqueriers, and join the 2 subqueries on CID. Define your ORDER BY in the parent query instead of in a subquery.
SELECT
sub1.CID,
sub1.TotalOrders,
sub2.TotalDollarAmount
FROM
(
SELECT
CID,
Count(Order.OrderID) AS TotalOrders
FROM [Order]
GROUP BY CID
) AS sub1
INNER JOIN
(
SELECT
CID,
Sum(OrderItem.Quantity*OrderItem.SalePrice)
AS TotalDollarAmount
FROM OrderItem INNER JOIN [Order]
ON OrderItem.OrderID = [Order].OrderID
GROUP BY CID
) AS sub2
ON sub1.CID = sub2.CID
ORDER BY sub1.TotalOrders DESC;

How to get last children records with parent record from database

I have database with two tables:
Customers (Id PK, LastName)
and
Orders (Id PK, CustomerId FK, ProductName, Price, etc.)
I want to retrieve only customer' last orders details together with customer name.
I use .NET L2SQL but I think it's SQL question more than LINQ question so I post here SQL query I tried:
SELECT [t0].[LastName], (
SELECT [t2].[ProductName]
FROM (
SELECT TOP (1) [t1].[ProductName]
FROM [Orders] AS [t1]
WHERE [t1].[CustomerId] = [t0].[Id]
ORDER BY [t1].[Id] DESC
) AS [t2]
) AS [ProductName], (
SELECT [t4].[Price]
FROM (
SELECT TOP (1) [t3].[Price]
FROM [Orders] AS [t3]
WHERE [t3].[CustomerId] = [t0].[Id]
ORDER BY [t3].[Id] DESC
) AS [t4]
) AS [Price]
FROM [Customers] AS [t0]
Problem is that Orders has more columns (30) and with each column the query gets bigger and slower because I need to add next subqueries.
Is there any better way?
In SQL Server 2005 and above:
SELECT *
FROM (
SELECT o.*,
ROW_NUMBER() OVER (PARTITION BY c.id ORDER BY o.id DESC) rn
FROM customers c
LEFT JOIN
orders o
ON o.customerId = c.id
) q
WHERE rn = 1
or this:
SELECT *
FROM customers c
OUTER APPLY
(
SELECT TOP 1 *
FROM orders o
WHERE o.customerId = c.id
ORDER BY
o.id DESC
) o
In SQL Server 2000:
SELECT *
FROM customers с
LEFT JOIN
orders o
ON o.id =
(
SELECT TOP 1 id
FROM orders oi
WHERE oi.customerId = c.id
ORDER BY
oi.id DESC
)