I have a UIWebView
which loads a local index.html file.
However I have an external link in this html file that I'd like to open in Safari instead of internally in the UIWebView
.
Opening a link in safari from say a UIButton
was simple enough:
UIApplication.sharedApplication().openURL(NSURL(string: "http://www.stackoverflow.com"))
Opening the Instagram app from a external link also works like a charm.
<a href="instagram://media?id=434784289393782000_15903882">instagram://media?id=434784289393782000_15903882</a>
So my first though was to do something like this:
<a href="safari://stackoverflow.com">Open in Safari</a>
However that doesn't seem to work, then I read something about using webView:shouldStartLoadWithRequest:navigationType:
But everyone who's managed to open an external link in Safari is writing in Obj-C which I'm not too familiar with as I'm writing in Swift.
Update with Swift Code:
import UIKit
class AccessoriesViewController: UIViewController, UIWebViewDelegate {
@IBOutlet weak var webView:UIWebView!
override func viewDidLoad() {
super.viewDidLoad()
if let url = NSBundle.mainBundle().URLForResource("accessories", withExtension: "html") {
webView.loadRequest(NSURLRequest(URL: url))
}
}
override func preferredStatusBarStyle() -> UIStatusBarStyle {
return UIStatusBarStyle.LightContent;
}
func webView(webView: UIWebView, shouldStartLoadWithRequest request: NSURLRequest, navigationType: UIWebViewNavigationType) -> Bool {
if let url = request.URL where navigationType == UIWebViewNavigationType.LinkClicked {
UIApplication.sharedApplication().openURL(url)
return false
}
return true
}
override func didReceiveMemoryWarning() {
super.didReceiveMemoryWarning()
// Dispose of any resources that can be recreated.
}
}