如何确定当前的iPhone /设备型号？
有什么方法可以在Swift中获取设备型号名称（iPhone 4S，iPhone 5，iPhone 5S等）？
我知道有一个名为UIDevice.currentDevice().model的属性，但它只返回设备类型（iPod touch，iPhone，iPad，iPhone模拟器等）。
我也知道它可以在Objective-C中用这种方法轻松完成：
#import <sys/utsname.h>
struct utsname systemInfo;
uname(&systemInfo);
NSString* deviceModel = [NSString stringWithCString:systemInfo.machine
encoding:NSUTF8StringEncoding];
但我正在Swift开发我的iPhone应用程序，所以有人可以帮助我用Swift解决这个问题的等效方法？
我在UIDevice上做了这个“纯Swift”扩展。
此版本适用于Swift 2或更高版本如果您使用早期版本，请使用我的答案的旧版本。
如果你正在寻找一个更优雅的解决方案，你可以使用我在GitHub上发布的μ-framework DeviceKit （也可以通过CocoaPods，Carthage和Swift包管理器）。
代码如下：
import UIKit
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 "iPhone8,2": return "iPhone 6s Plus"
case "iPhone9,1", "iPhone9,3": return "iPhone 7"
case "iPhone9,2", "iPhone9,4": return "iPhone 7 Plus"
case "iPhone8,4": return "iPhone SE"
case "iPhone10,1", "iPhone10,4": return "iPhone 8"
case "iPhone10,2", "iPhone10,5": return "iPhone 8 Plus"
case "iPhone10,3", "iPhone10,6": return "iPhone X"
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 "iPad6,11", "iPad6,12": return "iPad 5"
case "iPad7,5", "iPad7,6": return "iPad 6"
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 "iPad6,3", "iPad6,4": return "iPad Pro 9.7 Inch"
case "iPad6,7", "iPad6,8": return "iPad Pro 12.9 Inch"
case "iPad7,1", "iPad7,2": return "iPad Pro 12.9 Inch 2. Generation"
case "iPad7,3", "iPad7,4": return "iPad Pro 10.5 Inch"
case "AppleTV5,3": return "Apple TV"
case "AppleTV6,2": return "Apple TV 4K"
case "AudioAccessory1,1": return "HomePod"
case "i386", "x86_64": return "Simulator"
default: return identifier
}
}
}
你这样称呼它：
// Swift 2
let modelName = UIDevice.currentDevice().modelName
// Swift 3
let modelName = UIDevice.current.modelName
这个Swift 3.0示例将当前设备模型作为enum常量返回（以避免直接比较字符串文字）。 枚举的原始值是一个String含有人可读iOS设备名称。 由于它是Swift，所以已识别设备的列表中只包含足够支持iOS版本的模型，其中包括Swift。 以下用法示例在此答案的末尾使用了实现：
switch UIDevice().type {
case .iPhone5:
print("No TouchID sensor")
case .iPhone5S:
fallthrough
case .iPhone6:
fallthrough
case .iPhone6plus:
fallthrough
case .iPad_Pro9_7:
fallthrough
case .iPad_Pro12_9:
fallthrough
case .iPhone7:
fallthrough
case .iPhone7plus:
print("Put your thumb on the " +
UIDevice().type.rawValue + " TouchID sensor")
case .unrecognized:
print("Device model unrecognized");
default:
print(UIDevice().type.rawValue + " not supported by this app");
}
您的应用程序应该保持最新的设备版本，当苹果为同一设备系列添加新的模型时。 例如，iPhone3,1 iPhone3,2 iPhone3,4都是“iPhone 4”。 避免编写不考虑新模型的代码，因此您的算法不会意外地无法配置或响应新设备。 您可以参考此维护的iOS设备型号列表，在战略时间更新您的应用。
iOS包括与设备无关的接口，用于检测硬件功能和参数（如屏幕尺寸）。 Apple提供的通用接口通常是动态调整应用程序行为以适应不同硬件的最安全，最佳支持的机制。 尽管如此，下面的代码对于原型设计，调试，测试或任何时间代码需要针对特定设备系列都很有用。 这种技术也可用于通过其公共/公认的名称来描述当前设备。
斯威夫特3
// 1. Declare outside class definition (or in its own file).
// 2. UIKit must be included in file where this code is added.
// 3. Extends UIDevice class, thus is available anywhere in app.
//
// Usage example:
//
// if UIDevice().type == .simulator {
// print("You're running on the simulator... boring!")
// } else {
// print("Wow! Running on a (UIDevice().type.rawValue)")
// }
import UIKit
public enum Model : String {
case simulator = "simulator/sandbox",
iPod1 = "iPod 1",
iPod2 = "iPod 2",
iPod3 = "iPod 3",
iPod4 = "iPod 4",
iPod5 = "iPod 5",
iPad2 = "iPad 2",
iPad3 = "iPad 3",
iPad4 = "iPad 4",
iPhone4 = "iPhone 4",
iPhone4S = "iPhone 4S",
iPhone5 = "iPhone 5",
iPhone5S = "iPhone 5S",
iPhone5C = "iPhone 5C",
iPadMini1 = "iPad Mini 1",
iPadMini2 = "iPad Mini 2",
iPadMini3 = "iPad Mini 3",
iPadAir1 = "iPad Air 1",
iPadAir2 = "iPad Air 2",
iPadPro9_7 = "iPad Pro 9.7"",
iPadPro9_7_cell = "iPad Pro 9.7" cellular",
iPadPro10_5 = "iPad Pro 10.5"",
iPadPro10_5_cell = "iPad Pro 10.5" cellular",
iPadPro12_9 = "iPad Pro 12.9"",
iPadPro12_9_cell = "iPad Pro 12.9" cellular",
iPhone6 = "iPhone 6",
iPhone6plus = "iPhone 6 Plus",
iPhone6S = "iPhone 6S",
iPhone6Splus = "iPhone 6S Plus",
iPhoneSE = "iPhone SE",
iPhone7 = "iPhone 7",
iPhone7plus = "iPhone 7 Plus",
iPhone8 = "iPhone 8",
iPhone8plus = "iPhone 8 Plus",
iPhoneX = "iPhone X",
unrecognized = "?unrecognized?"
}
public extension UIDevice {
public var type: Model {
var systemInfo = utsname()
uname(&systemInfo)
let modelCode = withUnsafePointer(to: &systemInfo.machine) {
$0.withMemoryRebound(to: CChar.self, capacity: 1) {
ptr in String.init(validatingUTF8: ptr)
}
}
var modelMap : [ String : Model ] = [
"i386" : .simulator,
"x86_64" : .simulator,
"iPod1,1" : .iPod1,
"iPod2,1" : .iPod2,
"iPod3,1" : .iPod3,
"iPod4,1" : .iPod4,
"iPod5,1" : .iPod5,
"iPad2,1" : .iPad2,
"iPad2,2" : .iPad2,
"iPad2,3" : .iPad2,
"iPad2,4" : .iPad2,
"iPad2,5" : .iPadMini1,
"iPad2,6" : .iPadMini1,
"iPad2,7" : .iPadMini1,
"iPhone3,1" : .iPhone4,
"iPhone3,2" : .iPhone4,
"iPhone3,3" : .iPhone4,
"iPhone4,1" : .iPhone4S,
"iPhone5,1" : .iPhone5,
"iPhone5,2" : .iPhone5,
"iPhone5,3" : .iPhone5C,
"iPhone5,4" : .iPhone5C,
"iPad3,1" : .iPad3,
"iPad3,2" : .iPad3,
"iPad3,3" : .iPad3,
"iPad3,4" : .iPad4,
"iPad3,5" : .iPad4,
"iPad3,6" : .iPad4,
"iPhone6,1" : .iPhone5S,
"iPhone6,2" : .iPhone5S,
"iPad4,1" : .iPadAir1,
"iPad4,2" : .iPadAir2,
"iPad4,4" : .iPadMini2,
"iPad4,5" : .iPadMini2,
"iPad4,6" : .iPadMini2,
"iPad4,7" : .iPadMini3,
"iPad4,8" : .iPadMini3,
"iPad4,9" : .iPadMini3,
"iPad6,3" : .iPadPro9_7,
"iPad6,11" : .iPadPro9_7,
"iPad6,4" : .iPadPro9_7_cell,
"iPad6,12" : .iPadPro9_7_cell,
"iPad6,7" : .iPadPro12_9,
"iPad6,8" : .iPadPro12_9_cell,
"iPad7,3" : .iPadPro10_5,
"iPad7,4" : .iPadPro10_5_cell,
"iPhone7,1" : .iPhone6plus,
"iPhone7,2" : .iPhone6,
"iPhone8,1" : .iPhone6S,
"iPhone8,2" : .iPhone6Splus,
"iPhone8,4" : .iPhoneSE,
"iPhone9,1" : .iPhone7,
"iPhone9,2" : .iPhone7plus,
"iPhone9,3" : .iPhone7,
"iPhone9,4" : .iPhone7plus,
"iPhone10,1" : .iPhone8,
"iPhone10,2" : .iPhone8plus,
"iPhone10,3" : .iPhoneX,
"iPhone10,6" : .iPhoneX
]
if let model = modelMap[String.init(validatingUTF8: modelCode!)!] {
return model
}
return Model.unrecognized
}
}
又一个/简单的替代方案（模型标识符参考https://www.theiphonewiki.com/wiki/Models）：
更新Swift 3/4的答案，包括字符串修整和模拟器支持：
func modelIdentifier() -> String {
if let simulatorModelIdentifier = ProcessInfo().environment["SIMULATOR_MODEL_IDENTIFIER"] { return simulatorModelIdentifier }
var sysinfo = utsname()
uname(&sysinfo) // ignore return value
return String(bytes: Data(bytes: &sysinfo.machine, count: Int(_SYS_NAMELEN)), encoding: .ascii)!.trimmingCharacters(in: .controlCharacters)
}
Swift 2的原始答案：
func modelIdentifier() -> String {
var sysinfo = utsname()
uname(&sysinfo) // ignore return value
return NSString(bytes: &sysinfo.machine, length: Int(_SYS_NAMELEN), encoding: NSASCIIStringEncoding)! as String
}