IFERROR Formula with Conditions on Display on Error

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I have an IFERROR formula that is working as is (below), but I need to add a conditional statement to determine what needs to display when there is an error.

If there is an error, AND Column RA Type = "RA Type", then display "Rent to Income Ratio". Otherwise, display "" (blank).

=IFERROR((Rent@row * 12) / [Prev Resident that had RA or Income]@row, "Rent to Income Ratio")

Ideas on how to do that?

Best Answer

  • Toufong Vang
    Toufong Vang ✭✭✭✭✭
    Answer ✓
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    Hi, Bethany.

    Try...

    =IFERROR((Rent@row * 12) / [Prev Resident that had RA or Income]@row, IF([Column RA Type]@row = "RA Type", "Rent to Income Ratio", ""))

    The syntax for IFERROR() is IFERROR( this_is_what_you_prefer_be_done, alternatively_do_this )

    Your alternate can be a value or another statement which, in this case, is:

    IF([Column RA Type]@row = "RA Type", "Rent to Income Ratio", "")

    Cheers!

Answers

  • Toufong Vang
    Toufong Vang ✭✭✭✭✭
    Answer ✓
    Options

    Hi, Bethany.

    Try...

    =IFERROR((Rent@row * 12) / [Prev Resident that had RA or Income]@row, IF([Column RA Type]@row = "RA Type", "Rent to Income Ratio", ""))

    The syntax for IFERROR() is IFERROR( this_is_what_you_prefer_be_done, alternatively_do_this )

    Your alternate can be a value or another statement which, in this case, is:

    IF([Column RA Type]@row = "RA Type", "Rent to Income Ratio", "")

    Cheers!

  • Bethany Garcia
    Bethany Garcia ✭✭✭✭
    Options

    @Toufong Vang that worked perfectly, thank you!

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