title: How to open an URL with parameter in Swift3
body: <p>if you click for example on someone’s Instagram button from my app 
   you  gonna open the link <code>instagram://app/</code> username in the profile</p>
  
  <p>let’s say a user put “mark20” as Instagram username
   You open the link <code>instagram://app/mark20/</code></p>
  
  <p>i want to open instagram with showing 'mark20' . currently it is open Instagram news feeds but not open 'mark20' profile . i want to button click open Instagram  user profile page </p>
  
  <p>For example </p>
  
  <p><a href="https://i.stack.imgur.com/mihwR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mihwR.png" alt="enter image description here"></a></p>
  
  <p>How can i do that </p>
  
  <pre><code>func shareToInstagram() {
  
          //let instagramURL = URL(string: "instagram://app/")
          //not working anymore 
          let instagramURL = URL(string: "instagram://app/mark20/")
          if (UIApplication.shared.canOpenURL(instagramURL! as URL)) {
  
              UIApplication.shared.open(instagramURL!, options: ["":""], completionHandler: nil)
  
          } else {
              print(" Instagram isn't installed ")
          }
      }
  </code></pre>