Re: Dsum return 0 instead of Null?
I don't think Joan meant DSum(...) * 1, I think she meant:
Nz(DSum(...)) * 1 converts Null to 0
This is because Nz(DSum(...)) converts Null to Empty
and Empty * 1 yields 0.
The proper way to do this is:
Nz(DSum(...), 0)
This makes use of the Nz function's optional second argument.
.
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