Many ways to do that in Swift:
We check the model below (we can only do a case sensitive search here):
class func isUserUsingAnIpad() -> Bool {
let deviceModel = UIDevice.currentDevice().model
let result: Bool = NSString(string: deviceModel).containsString("iPad")
return result
}
We check the model below (we can do a case sensitive/insensitive search here):
class func isUserUsingAnIpad() -> Bool {
let deviceModel = UIDevice.currentDevice().model
let deviceModelNumberOfCharacters: Int = count(deviceModel)
if deviceModel.rangeOfString("iPad",
options: NSStringCompareOptions.LiteralSearch,
range: Range<String.Index>(start: deviceModel.startIndex,
end: advance(deviceModel.startIndex, deviceModelNumberOfCharacters)),
locale: nil) != nil {
return true
} else {
return false
}
}
UIDevice.currentDevice().userInterfaceIdiom below only returns iPad if the app is for iPad or Universal. If it is an iPhone app being ran on an iPad then it won't. So you should instead check the model. :
class func isUserUsingAnIpad() -> Bool {
if UIDevice.currentDevice().userInterfaceIdiom == UIUserInterfaceIdiom.Pad {
return true
} else {
return false
}
}
This snippet below does not compile if the class does not inherit of an UIViewController, otherwise it works just fine. Regardless UI_USER_INTERFACE_IDIOM() only returns iPad if the app is for iPad or Universal. If it is an iPhone app being ran on an iPad then it won't. So you should instead check the model. :
class func isUserUsingAnIpad() -> Bool {
if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiom.Pad) {
return true
} else {
return false
}
}