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Help with query against AdventureWorks database - Customers purchasing more than 2

Need a hand with this query:

Return the average Sale Price for all customers who have purchased two or more products

Thanks!
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John500
Asked:
John500
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2 Solutions
 
cyberkiwiCommented:
Select AvgSalePrice = SUM(SumUnitSalePrice)/Sum(SumSaleQty)
From
(
select c.CustomerID, SUM(d.UnitPrice*d.OrderQty) as SumUnitSalePrice, SUM(d.OrderQty) as SumSaleQty
from SalesLT.Customer c
inner join SalesLT.SalesOrderHeader h on h.CustomerID=c.CustomerID
inner join SalesLT.SalesOrderDetail d on d.SalesOrderID=h.SalesOrderID
group by c.CustomerID
having COUNT(d.SalesOrderDetailID) > 1 -- 2 or more products
) X

Not sure how you define "average Sale Price", but here's one definition.
Weighted total sale price (unit * qty) divided by total qty.
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John500Author Commented:
Appreciate the feed back here.

I'm wondering if we have different versions of the database.  Supposedly, the latest version can be obtained here:

http://msftdbprodsamples.codeplex.com/releases/view/4004

For instance, I don't have the anything labled SalesLT.

If I change all the SalesLT's to Sales and run this, I get:

389.0395

How about you ?


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samijsrCommented:
cyberkiwi: solution having one bugs that it also calculte if same cutomer Purchased same item more than one, where as requirement is two or more items

Select SC.CustomerID,Avg(SumSalaesPrice)as 'AvgSlaesPrice' from Sales.Customer SC
Inner Join Sales.SalesOrderHeader SS on SC.CustomerID=SS.CustomerID
Inner Join
(Select SD.SalesOrderID,sum(LineTotal)as 'SumSalaesPrice' from Sales.SalesOrderDetail SD
Inner Join
(
Select SalesOrderID,count(SalesOrderID)as 'cnt'
from
(Select distinct SalesOrderID,ProductID from Sales.SalesOrderDetail)T1
Group by SalesOrderID having count(SalesOrderID)>1)T2
On SD.SalesOrderID=T2.SalesOrderID
group by SD.SalesOrderID)T3
on T3.SalesOrderID=SS.SalesOrderID
Group By SC.CustomerID
order by 1
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cyberkiwiCommented:
John,

Can you confirm the requirement?

> Return the average Sale Price for all customers who have purchased two or more products

1) purchased two or more products
Does that mean distinct productid?
Just quantity of more than 1 across all invoices?]
More than 1 invoice line across all invoices

2) average Sale Price for all customers
Is that one figure for a grand total average, the result is a single row, single column like the "389.0395" figure?
Is it for each customer, an average sale price?
Or do you actually want a per-product average, with each row in the output as one product?

3) average Sale Price, using sample of just 2 sales (1 sale of qty 10 x unit price $1 + 1 sale of 1 x $10)
Is that a weighted average = ( $20 / 11 )
or a simple average = ( $11 / 2 ) ?
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John500Author Commented:
Hey guys !

Thanks for the help here.  In answer to your questions, the goal is to determine the average price each individual is purchasing so long as that person has purchased at least two items.  Thus, an average can be determined.

So then, cyberkiwi's 3rd scenario is what we are after:

3) average Sale Price, using sample of just 2 sales (1 sale of qty 10 x unit price $1 + 1 sale of 1 x $10)
The weighted average = ( $20 / 11 )

The results should show two fields, the customerID and the average price for *everything* they have purchased.

I'm not certain but it may be most accurate or helpful to use OVER(PARTITION BY CustomerID) AS 'AVG_AMT' for this - yes/no ?  So in other words, the first SELECT statement would look something like this:

SELECT CustomerID, AVG(A.Amount_Column) OVER(PARTITION BY CustomerID) AS 'AVG_AMT'
...
...

Your thoughts?

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cyberkiwiCommented:
John,

That answers questions 2 and 3, but for 1, I am still a bit fuzzy.

select CustomerID, SumUnitSalePrice/SumSaleQty as [Average Sale Price]
from
(
select c.CustomerID, SUM(d.UnitPrice*d.OrderQty) as SumUnitSalePrice, SUM(d.OrderQty) as SumSaleQty
from Sales.Customer c
inner join Sales.SalesOrderHeader h on h.CustomerID=c.CustomerID
inner join Sales.SalesOrderDetail d on d.SalesOrderID=h.SalesOrderID
group by c.CustomerID
having COUNT(distinct d.ProductID) > 1 -- 2 or more products
) X

Tweak the having clause as required.  It will be one of

having COUNT(distinct d.ProductID) > 1
having SUM(d.OrderQty) > 1
having COUNT(d.SalesOrderDetailID) > 1
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John500Author Commented:
cyberkiwi,

That looks like the ticket.  The 'SalesOrderDetailID > 1' looks to be the answer as I see it.  That is, counting ALL the order details would appear to catch the whole shi-bang.  I may be wrong.

This is why I posted all of these questions.  I'm not familar with the AdventureWorks database and I need to provide answers in a short time frame.

Anyway, thanks both of you for the help on this!
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