query from a query or list with more than one result in colum

[bretdunalp]

First, I'm a total novice, I can usually find a workaround but it appears I'm stumped.

I have a query that sorts for an ID the result is a list of several other ID's Example Header- family ID Header- Subscriptions When I query for the family ID several subscriptions will come up. I need to take those subscription ID's and find those in one table/Query.

I tried in the criteria [Query]![AuthIDsHoldQuery]![Subscriptions] I get a pop up asking me to put in only one of the subscriptions but what I need is multiple.

Is there a formulaic solution (FYI I've even tried IN ([query]![Header])

[John_G]

How are you using this query? In a form/subform, or by itself?

Is the list of subscription ID's in one multivalue table field?

Can you give a bit more detail about your table structure?

[bretdunalp]

I will use it later in a form but right now just have the query bring up the information basically in a table there are multiple transactions for several subscriptions but only a single account has theses subscriptions. IE Joe Tompson is the account holder but he has several subscriptions 1234, 1235, 1137 etc in a large table are all the charges (with multiple accounts) that only list subscriptions associated with a change to the account over months. I can pull up all the subscription ID's associated with Joe then use that to filter the larger transaction table. my challenge is I'm getting in the final query sometimes multiple repeated transactions so I was wondering if there was a criteria way of doing this.

[John_G]

...I can pull up all the subscription ID's associated with Joe...

How do you do that? Where are these ID's stored - in which table and field? If they are in the account holder table, that is not the correct place for them.

What you need to have is the Account Holder ID as a Foreign Key (FK) in the Subscription table - that is the normal design for a one-to-many relationship, and it makes queries such as yours much easier.

[bretdunalp]

Ok after pulling my hair out it appears if I went into properties in the Queries and clicked unique values it solved it thanks for the Help.