Formula reference in IF statement unparseable or invalid operation.

mccoy_FSI

I have a column called [Limitation]. If that column has the value B1.3 in it, the [Reference] column should return "3", otherwise "NA".

I am using =IF([Limitation]:[Limitation] = "B1.3", "3", "NA")

It seems like it is right, but it's unparseable. Any ideas?

I will add that =IF([Limitation]1="B1.3","3","NA") works just fine. But I need it to be a column formula. And the option to convert to column formula does not work (error= "syntax is quite right")

Andrée Starå

Hi @mccoy_FSI

I hope you're well and safe!

Try something like this.

=IF(CONTAINS("B1.3", Limitation:Limitation), "3", "NA")

Did that work/help?

I hope that helps!

Be safe, and have a fantastic week!

Best,

Andrée Starå | Workflow Consultant / CEO @ WORK BOLD

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mccoy_FSI

It just returns 3 in every row. Which is odd because some of those rows definitely do not contain B1.3. Though they do contain B1.1

Andrée Starå

@mccoy_FSI

Not sure I follow!

Do you want to check the whole column or only the row?

Can you elaborate?

mccoy_FSI

If the row has a B1.3 in the Limitation column, the reference column should be "3" in that row. Need the formula to apply new entries, hence why I think I need a column formula.

So row 1 may have B1.3 in the limitation column, therefore that rows reference column should be 3. Row 2 may have C in the limitation column, therefore that rows reference column should be NA.

Andrée Starå

@mccoy_FSI

Ok. Then this should work!

=IF(Limitation@row = "B1.3", 3, "N/A")

Did it?

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mccoy_FSI

Yes. Yes it did work. Thank you very much!

Andrée Starå

@mccoy_FSI

Excellent!

Happy to help!

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mccoy_FSI

Is this going to work with nested statements? For example, if Limitation column is B1.3 then reference column is 3 but if Limitation column is C then reference column is 1.

Update: Disregard that, yes nesting is pretty easy with that formula I got it to work how I need. Thanks again!