Reputation: 1169
On iOS 8 Apple gave us the possibility to go to the App Settings right from our app, using the Constant UIApplicationOpenSettingsURLString
UIApplication.sharedApplication().openURL(NSURL.URLWithString(UIApplicationOpenSettingsURLString))
There's a code to test if this Constant exists on iOS 7, but it uses ObjC and pointes. Apple did this on their code: https://developer.apple.com/library/ios/samplecode/AppPrefs/Listings/RootViewController_m.html
How can I make something like this using Swift?
Upvotes: 12
Views: 11129
Reputation: 25887
You could use #available:
if #available(iOS 8.0, *) {
UIApplication.sharedApplication().openURL(NSURL(string: UIApplicationOpenSettingsURLString)!);
} else {
// Fallback on earlier versions
};
Upvotes: 1
Reputation: 17081
In Swift:
switch UIDevice.currentDevice().systemVersion.compare("8.0.0", options: NSStringCompareOptions.NumericSearch) {
case .OrderedSame, .OrderedDescending:
UIApplication.sharedApplication().openURL(NSURL.URLWithString(UIApplicationOpenSettingsURLString))
case .OrderedAscending:
//Do Nothing.
}
In Objective-C:
if (&UIApplicationOpenSettingsURLString != NULL) {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];
}
Upvotes: 22
Reputation: 1169
It's not the same as checking for a Constant, but if you know the API version that contains the Constant (in my case, iOS 8), there are some techniques: http://nshipster.com/swift-system-version-checking/
One of them is:
switch UIDevice.currentDevice().systemVersion.compare("8.0.0", options: NSStringCompareOptions.NumericSearch) {
case .OrderedSame, .OrderedDescending:
println("iOS >= 8.0")
case .OrderedAscending:
println("iOS < 8.0")
}
Upvotes: 1