On the settings page, I want to include three links to
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.
This question is related to
ios
url
mobile-safari
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
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)
}
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 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)
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)
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)
}
Try this:
NSString *URL = @"xyz.com";
if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:URL]])
{
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:URL]];
}
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)
}
The non deprecated Objective-C version would be:
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://apple.com"] 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.)
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: [:])
}
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];
Source: Stackoverflow.com