Checking For Numeric Before Computing?

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PeteCresswell - 11 Jul 2007 14:39 GMT

Seems like:

=IF(ISNUMBER(R8C15),IF(R8C15 > 0, R8C15/R1C17,Null),Null)

Should return Null if R8C15 contains "xxx".

But it returns "#NAME?" instead.

Can anybody see what I'm doing wrong?

(I'm pushing that "IF" statement into the cell's .FormulaR1C1 via VBA
code)

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PeteCresswell - 11 Jul 2007 14:41 GMT

> Seems like:
>
[quoted text clipped - 8 lines]
> (I'm pushing that "IF" statement into the cell's .FormulaR1C1 via VBA
> code)

PS: It returns the expected ".1" if R1C17=10 and R8C15=1

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Ron Rosenfeld - 11 Jul 2007 14:57 GMT

>Seems like:
>
[quoted text clipped - 8 lines]
>(I'm pushing that "IF" statement into the cell's .FormulaR1C1 via VBA
>code)

Null is not a keyword. It needs to be in quotes.

=IF(ISNUMBER(R8C15),IF(R8C15 > 0, R8C15/R1C17,"Null"),"Null")

--ron

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Dave Peterson - 11 Jul 2007 16:26 GMT

or maybe you wanted an empty string returned so the cell looked blank.

=IF(ISNUMBER(R8C15),IF(R8C15 > 0, R8C15/R1C17,""),"")

But don't expect that cell to be really empty--it has a formula in it.

> Seems like:
>
[quoted text clipped - 8 lines]
> (I'm pushing that "IF" statement into the cell's .FormulaR1C1 via VBA
> code)

Signature

Dave Peterson

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PeteCresswell - 11 Jul 2007 19:35 GMT

> or maybe you wanted an empty string returned so the cell looked blank.

That's probably what I should be doing anyhow...

Thanks Ron.

Thanks Dave.

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