Hi,
I am trying to find out key column value and if it contain {3,9}, I want to show the specific column. But, I am encountering below error. Can someone advice what needs to be done.
Thanks in advance.
Could not display interface. Please check definition and inputs. Interface Definition: Expression evaluation error at function a!gridField [line 2]: Invalid types, can only act on data of the same type (Number (Integer), Record Field)
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Reference the record field like:
fv!row[recordtype!recordName.field]
I am getting the below error when I try use like fv!row[recordtype!recordName.field]
Could not display interface. Please check definition and inputs. Interface Definition: Expression evaluation error at function a!gridField [line 6]: Could not find variable 'fv!row'
fv!row is only available inside the "value" parameter. Making a whole column visible depending on the value of an individual cell also does not make that much sense.
I didn't had a clear look first, as Stefan mentioned this do not makes much sense but still if you have a requirement to do so then put your gridColumn showWhen condition in recordActionField showWhen paramater.
If we add showWhen condition inside the "recordActionField", then the column remain as-is without action button. But, I need to hide entire column.
Is there a way for this.
Query the same data which is in data parameter of gridField in a localVariable using queryRecordType() and then apply showWhen on gridColumns like:
contains({3,9}, local!data[recordType.fieldName])
Thank you so much.