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=IF Function, I think

Occasional Contributor

=IF Function, I think

=COUNTIF(C3,">"""),IF(AND(X3=FALSE,Y3=FALSE,Z3=FALSE,AA3=FALSE,AB3=FALSE,AC3=FALSE),"Risk",IF(AND(X3=TRUE,Y3=TRUE,Z3=TRUE,AA3=TRUE,AB3=TRUE,AC3=TRUE),"No Risk",""))

The idea of which is, if C3 is greater than null, then we need to look at the next part, so if X3,Y3,Z3,AA3,AB3,AC3 are all false then it should result in this cell showing Risk, but if any of the cells X3,Y3,Z3,AA3,AB3,AC3 are true then No Risk would show.

I know that the two parts work separately, but I just cannot find a way to join the Countif and the IF functions.

13 Replies

Re: =IF Function, I think

Hello Andy,
I guess that this is what you need:
=IF(C3=“”+X3+Y3+Z3+AA3+AB3+AC3,
“No Risk”,
“Risk”)

Re: =IF Function, I think

Hi Twifoo

Many thanks for your help, much simpler than my way, but comes back with a #value!

I am guessing that, that because although looking at X3-AC3, it is not looking for a True or False statement against them.  In this case, True would mean that if any one of the X3 to AC3 are true, then that would =No Risk, and a false to X3 to AC3 would = Risk

Re: =IF Function, I think

Try this:
=IF(ISBLANK(C3)+X3+Y3+Z3+AA3+AB3+AC3,
“No Risk”,
Risk”)

Re: =IF Function, I think

The last argument should be “Risk”

Re: =IF Function, I think

Given that X3:AC3 is a contiguous range, the formula can be reduced to this:
=IF(ISBLANK(C3)+SUMPRODUCT(—X3:AC3),
“No Risk”,
“Risk”)

Re: =IF Function, I think

Thanks for you continued help, but still now working, anything you can think of?

Re: =IF Function, I think

If the formula returns your desired result, then that should be it.

Re: =IF Function, I think

Twifoo, thank you, but please replace now with NOT, sorry my mistake

Re: =IF Function, I think

What result does the formula return?
Solution

Re: =IF Function, I think

Try this:
=IF((C3="")+SUMPRODUCT(--X3:AC3),
“No Risk”,
“Risk”)

Re: =IF Function, I think

Works a treat, thank you very much for your time.

Re: =IF Function, I think

Works a treat, thank you for both your time and effort

Re: =IF Function, I think

You're very much welcome!
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