有没有办法在Swift中获得设备型号名称(iPhone 4S, iPhone 5, iPhone 5S等)?
我知道有一个名为UIDevice.currentDevice()的属性。模型，但它只返回设备类型(iPod touch, iPhone, iPad, iPhone模拟器等)。
我也知道在Objective-C中使用以下方法可以轻松完成:
#import <sys/utsname.h>
struct utsname systemInfo;
uname(&systemInfo);
NSString* deviceModel = [NSString stringWithCString:systemInfo.machine
encoding:NSUTF8StringEncoding];
但是我正在Swift中开发我的iPhone应用程序，所以有人可以帮助我用等效的方法在Swift中解决这个问题吗?
当前回答
我发现很多答案都用了字符串。我决定改变@HAS答案使用一个enum:
public enum Devices: String {
case IPodTouch5
case IPodTouch6
case IPhone4
case IPhone4S
case IPhone5
case IPhone5C
case IPhone5S
case IPhone6
case IPhone6Plus
case IPhone6S
case IPhone6SPlus
case IPhone7
case IPhone7Plus
case IPhoneSE
case IPad2
case IPad3
case IPad4
case IPadAir
case IPadAir2
case IPadMini
case IPadMini2
case IPadMini3
case IPadMini4
case IPadPro
case AppleTV
case Simulator
case Other
}
public extension UIDevice {
public var modelName: Devices {
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 Devices.IPodTouch5
case "iPod7,1": return Devices.IPodTouch6
case "iPhone3,1", "iPhone3,2", "iPhone3,3": return Devices.IPhone4
case "iPhone4,1": return Devices.IPhone4S
case "iPhone5,1", "iPhone5,2": return Devices.IPhone5
case "iPhone5,3", "iPhone5,4": return Devices.IPhone5C
case "iPhone6,1", "iPhone6,2": return Devices.IPhone5S
case "iPhone7,2": return Devices.IPhone6
case "iPhone7,1": return Devices.IPhone6Plus
case "iPhone8,1": return Devices.IPhone6S
case "iPhone8,2": return Devices.IPhone6SPlus
case "iPhone9,1", "iPhone9,3": return Devices.IPhone7
case "iPhone9,2", "iPhone9,4": return Devices.IPhone7Plus
case "iPhone8,4": return Devices.IPhoneSE
case "iPad2,1", "iPad2,2", "iPad2,3", "iPad2,4":return Devices.IPad2
case "iPad3,1", "iPad3,2", "iPad3,3": return Devices.IPad3
case "iPad3,4", "iPad3,5", "iPad3,6": return Devices.IPad4
case "iPad4,1", "iPad4,2", "iPad4,3": return Devices.IPadAir
case "iPad5,3", "iPad5,4": return Devices.IPadAir2
case "iPad2,5", "iPad2,6", "iPad2,7": return Devices.IPadMini
case "iPad4,4", "iPad4,5", "iPad4,6": return Devices.IPadMini2
case "iPad4,7", "iPad4,8", "iPad4,9": return Devices.IPadMini3
case "iPad5,1", "iPad5,2": return Devices.IPadMini4
case "iPad6,3", "iPad6,4", "iPad6,7", "iPad6,8":return Devices.IPadPro
case "AppleTV5,3": return Devices.AppleTV
case "i386", "x86_64": return Devices.Simulator
default: return Devices.Other
}
}
}
其他回答
我最近在开发这个swift包:
https://github.com/EmilioOjeda/Device
我认为它的主要优点是代码生成的。所以每当一个新的Xcode版本发布时，我所要做的就是运行一个脚本并更新swift包。
代码生成是如何工作的?
它从Xcode的数据库(每个平台)读取并解析设备的数据，以识别正在运行的设备，并为其提供信息。
这是用法……
import Device
// or => 'import iPhoneOSDevice' <= iPhoneOS-only device data
// or => 'import tvOSDevice' <= tvOS-only device data
// A representation of the running device - it could be either actual or simulated.
let device = Device.current
// It is 'true' when running on a physical device.
_ = device.isDevice
// It is 'true' when running on a simulator instance.
_ = device.isSimulator
// It is 'true' when the device does not match either simulators or physical devices.
_ = device.isUnknown
// The identifier for the device.
// i.e.: 'iPhone15,3'
let id = device.id
// The known/commercial name of the device.
// i.e.: 'iPhone 14 Pro Max'
let model = device.model
斯威夫特5
/// Obtain the machine hardware platform from the `uname()` unix command
///
/// Example of return values
/// - `"iPhone8,1"` = iPhone 6s
/// - `"iPad6,7"` = iPad Pro (12.9-inch)
static var unameMachine: String {
var utsnameInstance = utsname()
uname(&utsnameInstance)
let optionalString: String? = withUnsafePointer(to: &utsnameInstance.machine) {
$0.withMemoryRebound(to: CChar.self, capacity: 1) {
ptr in String.init(validatingUTF8: ptr)
}
}
return optionalString ?? "N/A"
}
还有另一个简单的替代方案(模型标识符参考可在https://www.theiphonewiki.com/wiki/Models):找到)
Swift 3/4/5的更新答案，包括字符串修剪和模拟器支持:
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)
}
我的简单解决方案是按设备分组，支持新设备iPhone 8和iPhone X在Swift 3:
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 "iPhone3,1", "iPhone3,2", "iPhone3,3", "iPhone4,1":
return "iPhone 4"
case "iPhone5,1", "iPhone5,2", "iPhone5,3", "iPhone5,4", "iPhone6,1", "iPhone6,2", "iPhone8,4":
return "iPhone 5"
case "iPhone7,2", "iPhone8,1", "iPhone9,1", "iPhone9,3", "iPhone10,1", "iPhone10,4":
return "iPhone 6,7,8"
case "iPhone7,1", "iPhone8,2", "iPhone9,2", "iPhone9,4", "iPhone10,2", "iPhone10,5":
return "iPhone Plus"
case "iPhone10,3", "iPhone10,6":
return "iPhone X"
case "i386", "x86_64":
return "Simulator"
default:
return identifier
}
}
}
和使用:
switch UIDevice.current.modelName {
case "iPhone 4":
case "iPhone 5":
case "iPhone 6,7,8":
case "iPhone Plus":
case "iPhone X":
case "Simulator":
default:
}
使用Swift 3 (Xcode 8.3)
func deviceName() -> String {
var systemInfo = utsname()
uname(&systemInfo)
let str = withUnsafePointer(to: &systemInfo.machine.0) { ptr in
return String(cString: ptr)
}
return str
}
注意:根据官方开发论坛的回答，以这种方式使用元组是安全的。大Int8元组的内存对齐方式将与大Int8数组相同。即:连续且无填充。
