I have an App that initially presents with a TableViewController where each cell leads to a ViewController simple.
However, each ViewController displays many URLs that are automatically recognized within TextViews and it would be very stressful to use a SafariVC process for each URL.
So how do I make the URLs of TextView open SafariViewController ?
I recommend using CSafariWebKit
In your file
import CSafariWebKit
Declare
optional func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath)
This function fires when a cell is selected. And then get your url in the cell of that IndexPath and call the SafariViewController as below.
let vc = SafariViewController(url: url, barTintColor: nil, tintColor: nil)
vc.presentSafari(fromViewController: self, whenDidFinish: nil)
Use the colors of your app for a better experience.