如何使用Swift 3检测iPod和iPhone设备？

我想检测的测试设备是iPod或iPhone。现在，我正在使用这个代码

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

    }
    else
    {
        debugPrint("ipod show")
    }

当我用iPhone7测试时，它显示的是iPod。所以，我想检测它是iPod还是iPhone
我不想安装任何额外的吊舱来实现这一点
我想用简单而简短的代码来实现这一点。

---

有人能帮我吗

您可以通过对UIDevice进行扩展来获得更好的性能，如：

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
    }
  }
}

用法：

UIDevice.current.modelName

这将以字符串形式返回设备模型

试着像这样使用它：

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

---

stacjoverflow中已经有一个相关主题是的。我看到了那个帖子。但是，它不包括iPod和iPhone。我如何检查它的可能重复？请你详细告诉我，兄弟？我也写了它的用法，只要调用UIDevice.modelName，它就会返回当前的设备型号名称，无论是iPhone、iPad、iPod Apple TV还是simulator simple。谢谢兄弟。我不太明白。就我而言，我想看看iPod或iPhone。那么，如果UIDevice.current.modelName==“iPod Touch 5”{print（“iPod Touch 5”）}或者{print（“Real Device”）}我应该这样检查吗？所以，我需要再次编写iPod Touch 6？您必须再次编写，然后尝试在开关盒中使用。如果您从另一个答案（可能是从这里？）复制代码，请确保您正确地对其进行了属性设置，并进行比较。如果您认为答案之前已经被询问和回答过，那么您可以简单地将其标记为重复答案，而不是重复现有答案。