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We need to run a query of parents that did NOT make a gift to last year's Annual Fund. How can I do that? Because when I put in total gifts = $0.00 nothing comes up, but Jane Smith clearly did not make a gift last year. I know there are others.
Many thanks!
I believe the following criteria would work (assuming you have a specific Fund set up for last year's Annual Fund):
I would suggest doing two queries and merging them together to get the results. The first query is all constituents (nothing in the criteria). The second query is constituents that have given to the Annual Fund (Gift amount > $0; Fund = 2007 Annual Fund). Then merge the query using Query 1 = All Constituents and Query 2 = Given to 2007 Annual Fund with the operator SUB. This should leave you with all constituents who did not give to the 2007 Annual Fund.
Ditto what Jeff stated. That's what we had to do to come up with a similar list. Worked fine.
Did you filter your total for only gifts to the annual fund last year? You also might try doing a count rather than a total.
But wouldn't a single query with:
Constituency-Code EQUAL "Parents" AND Fund-Description NOT EQUAL "2007 Annual Fund" AND Gift-Amount GREATER-THAN 0
give this same result with a lot less effort on the part of both the user and the computer?
Note that (I think) this query still won't show those constituents who PLEDGED to the "2007 Annual Fund", but who have had their pledges written off.
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No, it won't. If I consistently give $10 to the Annual Fund, then I'll have a gift to the "2006 Annual Fund" (<> "2007 Annual Fund) of $10 (> 0), so I will be selected by your query despite also having a gift of $10 to the "2007 Annual Fund."
The problem is the scope of the "Not". If you are familiar with logic, the question was something of the form ¬∃x f(x). What you did was to move the "not" inside of the "exists" statement to get ∃x ¬f(x), but those are demonstrabely not equivalent. In fact, ¬∃x f(x) can be shown to be equivalent to ∀x ¬f(x)
Drew
PS: I certainly hope those math characters come across correctly.