Re: [MS_AccessPros] FIELD JOIN IN QUERY

 

Thanks Duanne,

i am trying to solve a problem with some combinatorics element. There is no unique ID in the first table that equates to another ID in the second.

While the first table contains all the possibilities, my desired outcome is only those lines or records in the first table that have a record from the second table as a subset regardless of the order.

The challenge is how to join both tables to achieve this result when i run a query

Ade

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From: "Duane Hookom duanehookom@hotmail.com [MS_Access_Professionals]" <MS_Access_Professionals@yahoogroups.com>
To: "MS_Access_Professionals@yahoogroups.com" <MS_Access_Professionals@yahoogroups.com>
Sent: Monday, 30 October 2017, 14:38
Subject: Re: [MS_AccessPros] FIELD JOIN IN QUERY

 

Ade,

It seems your tables aren't normalized correctly. Can you provide more information about what types of values you are storing? I am picturing something like storing EmployeeIDs  (40 different employees) in any of the fields with headings/field names that equate to positions or roles. If this is somewhat like you system, rather than 5 fields, there should be 5 records.

How would you join? Can you provide some sample records and desired output?

Regards,

Duane Hookom

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From: MS_Access_Professionals@yahoogroups.com <MS_Access_Professionals@yahoogroups.com> on behalf of Adeboyejo Oyenuga aoye_99@yahoo.co.uk [MS_Access_Professionals] <MS_Access_Professionals@yahoogroups.com>
Sent: Monday, October 30, 2017 6:02 AM
To: ms_access_professionals@yahoogroups.com
Subject: [MS_AccessPros] FIELD JOIN IN QUERY

 

I am looking for help to create a join between two tables. Fields 1-5 in the first table can take on about 40 different unique values, while Fields 1-3 in the second table can also take on the same number of unique values. Fields 1 to 3 can each equate to any of the values in Fields 1-5 in the 1st table.

Thanks.

Ade

TABLE1
FIELD1	FIELD2	FIELD3	FIELD4	FIELD5
TABLE2
FIELD1	FIELD2	FIELD3

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Posted by: Adeboyejo Oyenuga <aoye_99@yahoo.co.uk>

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