How to detect iPod and iPhone device with Swift 3?

Question

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?

There is a relative topic already in stacjoverflow [This link](https://stackoverflow.com/questions/24059327/detect-current-device-with-ui-user-interface-idiom-in-swift) — A.Patsogiannis, Jun 15 '17 at 04:41
Yes. I saw that post. But, it does not include differentiation iPod and iPhone. — May Phyu, Jun 15 '17 at 04:43
Possible duplicate of [iOS: How to determine the current iPhone/device model in Swift?](https://stackoverflow.com/questions/26028918/ios-how-to-determine-the-current-iphone-device-model-in-swift) — bestiosdeveloper, Jun 15 '17 at 06:40

bestiosdeveloper · Accepted Answer · 2017-06-15T04:51:01.947

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")
 }

I've also written the usage of it just call UIDevice.modelName and it'll return you the current device model name weather it is iPhone, iPad, iPod Apple TV or simulator simple. — bestiosdeveloper, Jun 15 '17 at 04:49
Thanks bro. I don't understand clearly. For my case, I would like to check iPod or iPhone. So, if UIDevice.current.modelName == "iPod Touch 5" { print("iPod Touch 5") } else { print("Real Device") } Should I check like this? So, I need to write iPod Touch 6 again? — May Phyu, Jun 15 '17 at 05:01
If you copy code from another answer (perhaps from here https://stackoverflow.com/a/26962452/1187415 ?) then make sure that you attribute it correctly, compare https://stackoverflow.com/help/referencing. If you think that the answer has been asked and answered before then you can simply flag it as a duplicate instead of repeating an existing answer. — Martin R, Jun 15 '17 at 06:05
@MartinR Thank you, as I was not aware about the flag option. — bestiosdeveloper, Jun 15 '17 at 06:21
You can also use like this: https://stackoverflow.com/questions/24059327/detect-current-device-with-ui-user-interface-idiom-in-swift — anas.p, Oct 21 '17 at 04:21

How to detect iPod and iPhone device with Swift 3?

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