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MCaliebe
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Count values in a query

I have a query in which I query ten tables for cross reference information.

Each table has
Item_nr
Vendor_item_nr
Price

My master table has
Item_nr
30 other fields

My query selects Vendor_item_nr from each table where Item_Nr=tbl_master!item_nr

My results
Item_nr, tbl_1!Vendor_item_nr, tbl_2!Vendor_item_nr, tbl_3!Vendor_item_nr...ect

How can I count the number of Vendor_item_nr results returned for each Item_nr?
Microsoft Access

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Dale Fye

8/22/2022 - Mon
SOLUTION
jerryb30

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Dale Fye

You need to  get rid of the 10 separate (Vendor) tables and normalize that data into a single table with fields:

VendorID
Item_nr
Vendor_item_nr
Price

With that type of a structure, you simply count the number of VendorID values for each Item_Nr.

You could accomplish the same thing with a normalizing query, but having separate tables for each vendor makes the entire database application far more complex than it needs to by.
jerryb30

Meant to say: Create a union query, and use as source of count query.
MCaliebe

ASKER
I was given each vendor as it's own spread sheet.  I admit I thought about putting them all together, however not all the vendors use a unique vendor_item_nr so I could have 3 vendors with unique Vendor_item_nr and 3 using the same. I needed to keep each record identified back to the vendor as well.  Some vendors have multiple prices as well.
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Dale Fye

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MCaliebe

ASKER
Although I'm certain Jerry's solution would work, I think the best solution for me is to do as fyed suggest and try to combine this data in a more data friendly format.  I'll probably have to ask this questions again, however at this time...I'll work with the data.
Dale Fye

You can use Jerry's union query to get the data into a normalized format I recommended.  That is where you need to start.  Then create a Make-Table query from the union query to create the new normalized table.  As long as you don't delete the original 10 vendor tables, you can tweak that union query until you get the structure the way you want it.