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.
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.
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)
openURL(_:)
was deprecated in iOS 10.0, instead you should call the instance method open(_:options:completionHandler:)
on UIApplication.
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.
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)
}
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.
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.)
openURL(_:)
was deprecated in iOS 10.0, instead you should call the instance method open(_:options:completionHandler:)
on UIApplication.
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];
text
here) won't be hard-coded and you need code like this.
The non deprecated Objective-C version would be:
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://apple.com"] options:@{} completionHandler:nil];
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
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)
}
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");
}
}];
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(:) 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
In SWIFT 3.0
if let url = URL(string: "https://www.google.com") {
UIApplication.shared.open(url, options: [:])
}
Try this:
NSString *URL = @"xyz.com";
if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:URL]])
{
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:URL]];
}
In Swift 1.2, try this:
let pth = "http://www.google.com"
if let url = NSURL(string: pth){
UIApplication.sharedApplication().openURL(url)
Swift 4 solution:
UIApplication.shared.open(NSURL(string:"http://yo.lo")! as URL, options: [String : Any](), completionHandler: nil)
Success story sharing