How to launch Safari and open URL from iOS app

Question

Here's what I did:

1. I created an IBAction in the header .h files as follows:

    - (IBAction)openDaleDietrichDotCom:(id)sender;
   

2. I added a UIButton on the Settings page containing the text that I want to link to.

3. I connected the button to IBAction in File Owner appropriately.

4. Then implement the following:

Objective-C

- (IBAction)openDaleDietrichDotCom:(id)sender {
    [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://www.daledietrich.com"]];
}

Swift

(IBAction in viewController, rather than header file)

if let link = URL(string: "https://yoursite.com") {
  UIApplication.shared.open(link)
}

Note that we do NOT need to escape string and/or address, like:

let myNormalString = "https://example.com";
let myEscapedString = myNormalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)!

In fact, escaping may cause opening to fail.

Swift Syntax:

UIApplication.sharedApplication().openURL(NSURL(string:"http://www.reddit.com/")!)

New Swift Syntax for iOS 9.3 and earlier

As of some new version of Swift (possibly swift 2?), UIApplication.sharedApplication() is now UIApplication.shared (making better use of computed properties I'm guessing). Additionally URL is no longer implicitly convertible to NSURL, must be explicitly converted with as!

UIApplication.sharedApplication.openURL(NSURL(string:"http://www.reddit.com/") as! URL)

New Swift Syntax as of iOS 10.0

The openURL method has been deprecated and replaced with a more versatile method which takes an options object and an asynchronous completion handler as of iOS 10.0

UIApplication.shared.open(NSURL(string:"http://www.reddit.com/")! as URL)

Here one check is required that the url going to be open is able to open by device or simulator or not. Because some times (majority in simulator) i found it causes crashes.

Objective-C

NSURL *url = [NSURL URLWithString:@"some url"];
if ([[UIApplication sharedApplication] canOpenURL:url]) {
   [[UIApplication sharedApplication] openURL:url];
}

Swift 2.0

let url : NSURL = NSURL(string: "some url")!
if UIApplication.sharedApplication().canOpenURL(url) {
     UIApplication.sharedApplication().openURL(url)
}

Swift 4.2

guard let url = URL(string: "some url") else {
    return
}
if UIApplication.shared.canOpenURL(url) {
    UIApplication.shared.open(url, options: [:], completionHandler: nil)
}

Take a look at the -openURL: method on UIApplication. It should allow you to pass an NSURL instance to the system, which will determine what app to open it in and launch that application. (Keep in mind you'll probably want to check -canOpenURL: first, just in case the URL can't be handled by apps currently installed on the system - though this is likely not a problem for plain http:// links.)

And, in case you're not sure if the supplied URL text has a scheme:

NSString* text = @"www.apple.com";
NSURL*    url  = [[NSURL alloc] initWithString:text];

if (url.scheme.length == 0)
{
    text = [@"http://" stringByAppendingString:text];
    url  = [[NSURL alloc] initWithString:text];
}

[[UIApplication sharedApplication] openURL:url];

The non deprecated Objective-C version would be:

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://apple.com"] options:@{} completionHandler:nil];