How can I launch Safari from an iPhone app?

Question

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

Hitormiss

Answer 1

should be the following :

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

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

Atrocious

Will this count towards your app's memory usage? Also, is there a good way to get back to your app (like the login feature in social networking sites)? – Varian

@Varian - Nothing is being hard alloc-ed here, so it's automatically set to autorelease. – Wording

@Varian my guess would be no as I assume the 'webview' is launched in the safari application so it would fall under that process – Atrocious

Hi is there any way I can open a link in safari web browser using javascript?? Like sending a link tag in my web view and opening a safari web browser – Crenelate

Can anyone answer the title question? namely, how does one open safari from an app (on a device)? Just safari app, not a certain address within safari app. Calling @"http://" opens a new blank tab in safari app. – Murage

Open url is useful because it returns NO if a parental restriction is turned on for Safari app. In one of my apps I found that one of my code paths introduced a bug by not handling this behaviour. – Chute

dupe of earlier 5/9/09 answer – Gourmont

@Barett: Not exactly actually, because that's a 9/21/09 answer – Shulock

IMO the API call is similar enough that this answer would have been better applied as an edit or comment on the prior answer. – Gourmont

Answer 2

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]);
}

Unhopedfor

Answer 3

you can open the url in safari with this:

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

Henriquez

Answer 4

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
}

Rhondarhondda

Answer 5

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")!)

Attah

Answer 6

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)
}

Lottie

While this code snippet may solve the question, including an explanation helps to improve the quality of your response. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion. – Conto

Answer 7

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)") })
    }
}

Handsel