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Problem: I want to show/hide a field based on the content of a field in an EXTENDED lookup column.

I have the following two lists:

LIST 1: "*Opgaver*"
 COLUMN: "*Opgave*"
 COLUMN: "*Område*"

LIST 2: "*Tidsregistrering*"
 COLUMN: "Aktivitet"
 COLUMN: "Underområde"
 COLUMN: "Opgave" (lookup)
 COLUMN: "Opgave: Område" (extended lookup based on lookup)

I use Power Automate to ensure that all involved columns are available in text format (just a sidenote to avoid confusion).

In list 2 I only want to show the field pertaining to "Underområde" if the field in column "Opgave: Område" in list 2 has a certain content. I have tried to use the following formular from Microsofts learning material:

=if(substring([$Opgave: Område],indexOf([$Opgave: Område],';#')+2,1000) == 'Toronto', 'true', 'false')

But this results in the error message "Use valid condition". I am unable to figure out if this is caused by a faulty reference ($Opgave: Område) or it is due to the fact that I simply cannot use an extended lookup column for conditional logic.

Please help - TIA

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  • Try using correct internal names of your columns in formula once: ganeshsanapblogs.wordpress.com/2023/04/17/…. Let me know if it works.
    – Ganesh Sanap - MVP
    Commented Oct 31, 2023 at 13:56
  • Extended lookup columns might not be officially supported in SharePoint conditional formulas, I need to do some tests to confirm this.
    – Ganesh Sanap - MVP
    Commented Oct 31, 2023 at 13:57
  • Thanks, @GaneshSanap-MVP. I retrieved the following internal name based on your blog: "Opgave%5Fx003a%5F%5Fx0020%5FOmr%5Fx00e5%5Fde". Not very handy :) But tried to use it and got the same error message ("Use valid condition").
    – rabbitco
    Commented Oct 31, 2023 at 14:10

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