CODEWITHSUNDEEP
iossafarilaunch
keuminotti
keuminotti

Reputation: 1453

How can I launch Safari from an iPhone app?

This might be a rather obvious question, but can you launch the Safari browser from an iPhone app?

Upvotes: 129

Views: 86690

Answers (7)

Jia Chen
Jia Chen

Reputation: 73

In swift 4 and 5, as OpenURL is depreciated, an easy way of doing it would be just

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

You can also use SafariServices. Something like a Safari window within your app.

import SafariServices

...

if let url = URL(string: "https://stackoverflow.com") {
    let safariViewController = SFSafariViewController(url: url)        
    self.present(safariViewController, animated: true)
}

Upvotes: 3

DZoki019
DZoki019

Reputation: 392

With iOS 10 we have one different method with completion handler:

ObjectiveC:

NSDictionary *options = [NSDictionary new];
//options can be empty
NSURL *url = [NSURL URLWithString:@"http://www.stackoverflow.com"];
[[UIApplication sharedApplication] openURL:url options:options completionHandler:^(BOOL success){
}];

Swift:

let url = URL(string: "http://www.stackoverflow.com")
UIApplication.shared.open(url, options: [:]) { (success) in
}

Upvotes: 5

lisp-ceo
lisp-ceo

Reputation: 119

In Swift 3.0, you can use this class to help you communicate with. The framework maintainers have deprecated or removed previous answers.

import UIKit

class InterAppCommunication {
    static func openURI(_ URI: String) {
        UIApplication.shared.open(URL(string: URI)!, options: [:], completionHandler: { (succ: Bool) in print("Complete! Success? \(succ)") })
    }
}

Upvotes: 1

Jens Peter
Jens Peter

Reputation: 765

Maybe someone can use the Swift version:

In swift 2.2:

UIApplication.sharedApplication().openURL(NSURL(string: "https://www.google.com")!)

And 3.0:

UIApplication.shared().openURL(URL(string: "https://www.google.com")!)

Upvotes: 4

surtyaar
surtyaar

Reputation: 2542

should be the following : 

NSURL *url = [NSURL URLWithString:@"http://www.stackoverflow.com"];

if (![[UIApplication sharedApplication] openURL:url]) {
    NSLog(@"%@%@",@"Failed to open url:",[url description]);
}

Upvotes: 200

Brad The App Guy
Brad The App Guy

Reputation: 16275

UIApplication has a method called openURL:

example:

NSURL *url = [NSURL URLWithString:@"http://www.stackoverflow.com"];

if (![[UIApplication sharedApplication] openURL:url]) {
  NSLog(@"%@%@",@"Failed to open url:",[url description]);
}

Upvotes: 53

Dhaval Parmar
Dhaval Parmar

Reputation: 293

you can open the url in safari with this:

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"https://www.google.com"]];

Upvotes: 16

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