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How to launch Safari and open URL from iOS app

On the settings page, I want to include three links to

My app support site

YouTube app tutorial

My primary site (ie: linked to a 'Created by Dale Dietrich' label.)

I've searched this site and the web and my documentation and I've found nothing that is obvious.

NOTE: I don't want to open web pages within my app. I just want to send the link to Safari and that link be open there. I've seen a number of apps doing the same thing in their Settings page, so it must be possible.

Same issue i am facing in Hybrid development using Ionic Cordova app

T
Top-Master

Here's what I did:

I created an IBAction in the header .h files as follows: - (IBAction)openDaleDietrichDotCom:(id)sender; I added a UIButton on the Settings page containing the text that I want to link to. I connected the button to IBAction in File Owner appropriately. 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.


But this is not 100%, because if someone use a jailbreaked iOS and use Chrome or something as default browser, then this will open that, not Safari
I feel like if someone's gone to the effort of jailbreaking their phone to make Chrome their default browser, honouring that setting is probably the ideal behaviour.
I understood this but if I am opening a website and user is surfing the website now if he stops by a particular page then can I get the current webpage link in my code?
Above method is deprecated now use following. [[UIApplication sharedApplication] openURL:[NSURL URLWithString:self.advertisement.url] options:@{} completionHandler:^(BOOL success) { }];
@Laszlo Also, in settings we can disable Safari in Restrictions. I coded an app with as URL schemes http and https to open other browsers when an http or https URL is opened and Safari is disabled. So, even without Jailbreak, we can technically change the default web browser.
M
Michael Peterson

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)

Swift versions of answers are extremely useful to Swift developers, it's a new language without a lot of documentation, esp. on solving real world iOS problems. In my case, autocomplete had chosen the "fileURLWithPath:" selector and I didn't realize that was why my URL was not opening, it was only by reading Dustin's answer that I saw that I should have been using the "string:" selector . So Dustin should be upvoted, not chastised.
@g_fred. Why would he not be able to include a Swift version?
openURL(_:) was deprecated in iOS 10.0, instead you should call the instance method open(_:options:completionHandler:) on UIApplication.
You can directly use URL instead of NSURL so you can save the cast.
I recommend never to force unwrap. if let url = NSURL(string:"http://www.reddit.com/") { UIApplication.shared.open(url) } will always serve your better. Might want to assert and error in the else statement to catch typos.
C
Chetan Prajapati

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)
}

You can always open HTTP / HTTPS. As of iOS 9, you need to whitelist any URL you wish to call canOpenURL on. I don't see why you wouldn't simply call openURL:url and if it fails, deal with the failure. canOpenURL is primarily used to detect the installation of an app without leaving the current app.
openURL(_:) was deprecated in iOS 10.0, instead you should call the instance method open(_:options:completionHandler:) on UIApplication.
T
Tim

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.)


Thanks Tim. You are right on point. I was adding my own answer as yours came in. Sorry about that. But at least there is now an easy how-to for those that follow me.
No worries - glad you found the solution!
openURL(_:) was deprecated in iOS 10.0, instead you should call the instance method open(_:options:completionHandler:) on UIApplication.
m
meaning-matters

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];

@Vlad This is just a hard-coded example. Often the url string (text here) won't be hard-coded and you need code like this.
@Vlad Were you frustrated, and if yes how/why? I assume people don't just copy past the code above.
meaning-matters It is up to you :)
A
Antzi

The non deprecated Objective-C version would be:

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

G
Greg T

Swift 3.0

if let url = URL(string: "https://www.reddit.com") {
    if #available(iOS 10.0, *) {
        UIApplication.shared.open(url, options: [:])
    } else {
        UIApplication.shared.openURL(url)
    }
}

This supports devices running older versions of iOS as well


B
Baher A

Swift 3 Solution with a Done button

Don't forget to import SafariServices

if let url = URL(string: "http://www.yoururl.com/") {
            let vc = SFSafariViewController(url: url, entersReaderIfAvailable: true)
            present(vc, animated: true)
        }

This is the perfect answer which works in both iOS and iOS Extensions. Other answers only work in apps and can't be used in extensions.
D
Diego Jiménez

Because this answer is deprecated since iOS 10.0, a better answer would be:

if #available(iOS 10.0, *) {
    UIApplication.shared.open(url, options: [:], completionHandler: nil)
}else{
    UIApplication.shared.openURL(url)
}

y en Objective-c

[[UIApplication sharedApplication] openURL:@"url string" options:@{} completionHandler:^(BOOL success) {
        if (success) {
            NSLog(@"Opened url");
        }
    }];

Z
Zgpeace

Swift 5:

func open(scheme: String) {
   if let url = URL(string: scheme) {
      if #available(iOS 10, *) {
         UIApplication.shared.open(url, options: [:],
           completionHandler: {
               (success) in
                  print("Open \(scheme): \(success)")
           })
     } else {
         let success = UIApplication.shared.openURL(url)
         print("Open \(scheme): \(success)")
     }
   }
 }

Usage:

open(scheme: "http://www.bing.com")

Reference:

OpenURL in iOS10


R
Ryan H.

openURL(:) was deprecated in iOS 10.0, instead you should use the following instance method on UIApplication: open(:options:completionHandler:)

Example using Swift
This will open "https://apple.com" in Safari.

if let url = URL(string: "https://apple.com") {
    if UIApplication.shared.canOpenURL(url) {
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
    }
}

https://developer.apple.com/reference/uikit/uiapplication/1648685-open


O
OurangZeb Khan

In SWIFT 3.0

               if let url = URL(string: "https://www.google.com") {
                 UIApplication.shared.open(url, options: [:])
               }

a
arghtype

Try this:

NSString *URL = @"xyz.com";
if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:URL]])
{
     [[UIApplication sharedApplication] openURL:[NSURL URLWithString:URL]];
}

a
aircraft

In Swift 1.2, try this:

let pth = "http://www.google.com"
    if let url = NSURL(string: pth){
        UIApplication.sharedApplication().openURL(url)

plz tell me what is the new in code which you have written and you marked me down vote
K
Kevin ABRIOUX

Swift 4 solution:

UIApplication.shared.open(NSURL(string:"http://yo.lo")! as URL, options: [String : Any](), completionHandler: nil)

A proper Swift 4 version would use URL and not NSURL. And casting NSURL as URL is not the same. See stackoverflow.com/a/37812485/2227743 for an explanation. Also, force unwrapping is bad.