IF Formula with Multiple Criteria

Hi all,

Working on an IF formula with 3 conditions.

IF Value = 0, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use."

IF Value, 2-36, "Low probability of Risk: No Computer Validation testing required."

IF Value, =>36, "Medium probability of Risk: Computer Validation testing required."

I've used the following formulas, but I'm not having luck with the text appearing if the values are 2-36 or =>36.

#INCORRECT ARGUEMENT SET Error appears.

=IF([RPN (Severity x Probability x Detection)]@row = 1, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use.", IF([RPN (Severity x Probability x Detection)]@row, >=36, "Medium probability of Risk: Computer Validation testing required.", "Low probability of Risk: No Computer Validation testing required."))

=IF([RPN (Severity x Probability x Detection)]@row = 1, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use.", IF([RPN (Severity x Probability x Detection)]@row, <36, "Low probability of Risk: No Computer Validation testing required.", IF([RPN (Severity x Probability x Detection)]@row, >=36, "Medium probability of Risk: Computer Validation testing required.")))

Assuming I can also using an IFERROR formula if the value is 0 also?

Thanks,

Jack

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Accepted answers

SolutionSal

It looks like you have an extra comma in your logical_expression of a few IF statements. For example:

IF([RPN (Severity x Probability x Detection)]@row, >=36, "Medium probability of Risk: Computer Validation testing required.")))

Should look like this:

IF([RPN (Severity x Probability x Detection)]@row >=36, "Medium probability of Risk: Computer Validation testing required.")))

Here are your functions without the stray commas in the logical_expressions

=IF([RPN (Severity x Probability x Detection)]@row = 1, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use.", IF([RPN (Severity x Probability x Detection)]@row >=36, "Medium probability of Risk: Computer Validation testing required.", "Low probability of Risk: No Computer Validation testing required."))

=IF([RPN (Severity x Probability x Detection)]@row = 1, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use.", IF([RPN (Severity x Probability x Detection)]@row < 36, "Low probability of Risk: No Computer Validation testing required.", IF([RPN (Severity x Probability x Detection)]@row >=36, "Medium probability of Risk: Computer Validation testing required.")))

Hope this helps!

All comments

SolutionSal

It looks like you have an extra comma in your logical_expression of a few IF statements. For example:

IF([RPN (Severity x Probability x Detection)]@row, >=36, "Medium probability of Risk: Computer Validation testing required.")))

Should look like this:

IF([RPN (Severity x Probability x Detection)]@row >=36, "Medium probability of Risk: Computer Validation testing required.")))

Here are your functions without the stray commas in the logical_expressions

=IF([RPN (Severity x Probability x Detection)]@row = 1, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use.", IF([RPN (Severity x Probability x Detection)]@row < 36, "Low probability of Risk: No Computer Validation testing required.", IF([RPN (Severity x Probability x Detection)]@row >=36, "Medium probability of Risk: Computer Validation testing required.")))

Hope this helps!

Jack Parry

@SolutionSal This resolve the issue. Thank you!

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