CODEWITHSUNDEEP
iphoneswift3deviceipod
May Phyu
May Phyu

Reputation: 905

How to detect iPod and iPhone device with Swift 3?

I would like to detect test device is iPod or iPhone. Now, I'm using this code. 

if (UIDevice.current.userInterfaceIdiom == UIUserInterfaceIdiom.pad)
    {    debugPrint("ipad show")

    }
    else
    {
        debugPrint("ipod show")
    }

When I test with iPhone 7, it shows iPod. So, I would like to detect whether it is iPod or iPhone.
I don't want to install any additional pod to achieve this.
I would like to implement this with simply and short code.
Can anyone help me please?

Upvotes: 2

Views: 6716

Answers (1)

Pramod Kumar
Pramod Kumar

Reputation: 2652

You can get it better by making an extension for UIDevice like: 

public extension UIDevice {

var modelName: String {
    var systemInfo = utsname()
    uname(&systemInfo)
    let machineMirror = Mirror(reflecting: systemInfo.machine)
    let identifier = machineMirror.children.reduce("") { identifier, element in
        guard let value = element.value as? Int8, value != 0 else { return identifier }
        return identifier + String(UnicodeScalar(UInt8(value)))
    }

    switch identifier {
    case "iPod5,1":                                 return "iPod Touch 5"
    case "iPod7,1":                                 return "iPod Touch 6"
    case "iPhone3,1", "iPhone3,2", "iPhone3,3":     return "iPhone 4"
    case "iPhone4,1":                               return "iPhone 4s"
    case "iPhone5,1", "iPhone5,2":                  return "iPhone 5"
    case "iPhone5,3", "iPhone5,4":                  return "iPhone 5c"
    case "iPhone6,1", "iPhone6,2":                  return "iPhone 5s"
    case "iPhone7,2":                               return "iPhone 6"
    case "iPhone7,1":                               return "iPhone 6 Plus"
    case "iPhone8,1":                               return "iPhone 6s"
    case "iPhone9,1", "iPhone9,3":                  return "iPhone 7"
    case "iPhone9,2", "iPhone9,4":                  return "iPhone 7 Plus"
    case "i386", "x86_64":                          return "Simulator"
    case "iPad2,1", "iPad2,2", "iPad2,3", "iPad2,4":return "iPad 2"
    case "iPad3,1", "iPad3,2", "iPad3,3":           return "iPad 3"
    case "iPad3,4", "iPad3,5", "iPad3,6":           return "iPad 4"
    case "iPad4,1", "iPad4,2", "iPad4,3":           return "iPad Air"
    case "iPad5,3", "iPad5,4":                      return "iPad Air 2"
    case "iPad2,5", "iPad2,6", "iPad2,7":           return "iPad Mini"
    case "iPad4,4", "iPad4,5", "iPad4,6":           return "iPad Mini 2"
    case "iPad4,7", "iPad4,8", "iPad4,9":           return "iPad Mini 3"
    case "iPad5,1", "iPad5,2":                      return "iPad Mini 4"
    case "iPad6,7", "iPad6,8":                      return "iPad Pro"
    case "AppleTV5,3":                              return "Apple TV"
    case "i386", "x86_64":                          return "Simulator"
    default:                                        return identifier
    }
  }
}

Usage: UIDevice.current.modelName This will return the device model in string.

Try to use it like:

 if UIDevice.current.modelName == "Simulator" {
    print("Simulator")
 }
 else {
    print("Real Device")
 }

Upvotes: 6

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