canOpenURL: failed for URL: “instagram://app” - error: “This app is not allowed to query for scheme instagram”

Question

This is the code I use:

let instagramURL = NSURL(string: \"instagram://app\")
if UIApplication.shared.canOpenURL(instagramURL! as URL) {
  //Code
} else {
           


        

Answer 1

For those who try to open the app using custom URL Scheme (assume that app FirstApp opens SecondApp):

• In FirstApp add LSApplicationQueriesSchemes with URL Scheme to Info.plist like this:

• In SecondApp register new URL Scheme in URL Types like this: