I’ve read this by @Fabian but I’m not sure its the same issue
I’ve created a TEXT type UserSetting column called User_Type which gets the user permission level with the following set as the App formula
ANY(Select(Staff[Type],[Login Email]=UserEmail()))
I then wanted to use this to create a Show_If for a column with something like these:
UserSettings(“User_Type”)="Retailer"
Errors after save with:
Column Name ‘Job Number’ in Schema ‘Bookings_Schema’ of Column Type ‘Text’ has an invalid Editable_If constraint ‘=UserSettings(“User_Type”)=“Retailer”’. Unable to find column ‘_THISUSER’, did you mean ‘_RowNumber’?
UserSettings(User_Type)="Retailer"
Doesn’t error but doesn’t = TRUE either
UserSettings([User_Type])="Retailer"
Errors in the formula checker with:
Function ‘USERSETTINGS’ should have exactly 1 parameters, the name of an option from the user settings
I then tried UserSettings(User_Type)="Retailer" in a virtual column, which didn’t return a result
So I changed the UserSetting from an App Formula to a Initial Value. Then everything worked
But I don’t want the users messing with this so I hid the UserSetting and now I’m back to:
Column Name ‘Job Number’ in Schema ‘Bookings_Schema’ of Column Type ‘Text’ has an invalid Editable_If constraint ‘=UserSettings(User_Type)=“Retailer”’. Unable to find column ‘_THISUSER’, did you mean ‘_RowNumber’?
Am I missing something? Is there some logic as to why UserSetting formulas have to be both an initial value AND visible to work?
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