openurl

I just upgraded from XCode 6.4 to Xcode 7 GM and get started to change the code to be compliant with Swift 2. I could not come over the following errors. The project is a keyboard extension and the snippet is from the containing app. let s = "https://itunes.apple.com/" UIApplication.sharedApplication().openURL(NSURL(string : s)!) Error 1: 'sharedApplication()' is unavailable: Use view controller based solutions where appropriate instead. Error 2: 'openURL' is unavailable. This might be something to do with Xcode and AppDelegate, I might have screwed up my project. You can't access every API

In my iOS 4 application, I need to open a URL from the background, at a time specified from the user. However, for some reason, I cannot launch a URL from the background for some reason. Here is my code for opening a URL: if (![[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://www.stackoverflow.com"]]) { // the URL wasn't opened. we will ignore this for now. } That code is all being launched from a daemon thread created earlier. I have tested this code on the simulator, and the URL is not opened, and the method returns YES for some reason, but when I open my application

On iOS 8 beta 2 it should be possible to use openUrl from app extension as written into the release notes: however when I try to use this API (on Xcode 6 beta 2) I get the following error: Beta 2 really fixed this issue or not? Laurence Fan you may use this code: [self.extensionContext openURL:url completionHandler:^(BOOL success) { NSLog(@"fun=%s after completion. success=%d", __func__, success); }]; the API document: openURL:completionHandler: you could also refer to this question: openURL not work in Action Extension Accepted solution only works in Today extensions , a working solution in

I know that my app can open a particular URL in Safari by doing something like this: [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://www.example.com/"]]; but, is there any way to have my app switch over to Safari without opening a URL? I'd like to switch to Safari, but let it keep showing whatever page it had open the last time the user looked at it. Unfortunately no, unless you can figure out how to launch an app by bundle id in a non-jailbroken environment. Otherwise, if you are in a jailbroken environment, you can use the following to launch an app by its bundle id:

I have converted the code for making a phone call from Objective-C to Swift, but in Objective-C, we can set the type of the URL that we like to open (e.g. telephone, SMS, web) like this: @"tel:xx" @"mailto:info@example.es" @"http://stackoverflow.com" @"sms:768number" The code in Swift is: UIApplication.sharedApplication().openURL(NSURL(string : "9809088798") I read that have not released any scheme parameter for tel: , but I don't know if Swift can detect if the string is for making a phone call, sending email, or opening a website. Or may I write: (string : "tel//:9809088798") ? I am pretty

So apparently OpenURL has been deprecated in iOS 10. Does anyone have any documentation on why or can explain what to do next? I looked on the Apple site already and found a few things pertaining to OpenURL and this is what they say to use now: UIApplication.shared().open(url: URL, options: [String: AnyObject], completionHandler: ((Bool) -> Void)?) Does anyone have any evidence that this is the new way to use OpenURL in Swift 3.0? In addition what values are to be used in the options: and completionHandler: parameters respectively? You can also use conditional check if you are updating you app