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Sankar Ganesh How to display URL based on condition in detail page link
I have detail page URL button in which I have to open two URL based on some criteria. I tried using If condition but I am getting servlet keyword in the URL. can anyone suggest me how to remove servlet and open the correct link.
below is the URL format I tried.
In classic I am getting URL as /servlet/%2F00Oi0000006Cp9L
Can anyone help me to remove Servlet from URL. `
below is the URL format I tried.
{!IF( Campaign.UI_Theme__c = 'Theme4d', '/lightning/r/Report/00Oi0000006Cp9L/view?fv0={!Campaign.Name}', '/00Oi0000006Cp9L')}
In classic I am getting URL as /servlet/%2F00Oi0000006Cp9L
Can anyone help me to remove Servlet from URL. `