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
struct DeviceType {
static let IS_IPHONE_4_OR_LESS = UIDevice.current.userInterfaceIdiom == .phone && Constants.SCREEN_MAX_LENGTH < 568
static let IS_IPHONE_5 = UIDevice.current.userInterfaceIdiom == .phone && Constants.SCREEN_MAX_LENGTH == 568
static let IS_IPHONE_6 = UIDevice.current.userInterfaceIdiom == .phone && Constants.SCREEN_MAX_LENGTH == 667
static let IS_IPHONE_6P = UIDevice.current.userInterfaceIdiom == .phone && Constants.SCREEN_MAX_LENGTH == 736
static let IS_IPAD = UIDevice.current.userInterfaceIdiom == .pad && Constants.SCREEN_MAX_LENGTH == 1024
}
其他回答
这个Swift 3.0示例将当前设备模型作为枚举常量返回(以避免与字符串字面值直接比较)。enum的原始值是一个包含人类可读的iOS设备名称的字符串。因为它是Swift，所以认可的设备列表只包括最新的机型，足以支持包含Swift的iOS版本。下面的使用示例利用了这个答案末尾的实现:
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");
}
你的应用应该保持最新的新设备发布，以及当苹果为同一设备家族添加新型号时。例如，iPhone3、1 iPhone3、2 iPhone3、4都是“iPhone 4”。避免编写不考虑新型号的代码，这样你的算法就不会意外地配置失败或对新设备做出响应。你可以参考这个维护列表的iOS设备型号#更新你的应用程序在战略时刻。
iOS includes device-independent interfaces to detect hardware capabilities and parameters such as screen size. The generalized interfaces Apple provides are usually the safest, best supported mechanisms to dynamically adapt an app's behavior to different hardware. Nevertheless, the following code can be useful for prototyping, debugging, testing, or any time code needs to target a specific device family. This technique can also be useful to describe the current device by its common/publicly recognized name.
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",
iPhoneXS = "iPhone XS",
iPhoneXSmax = "iPhone XS Max",
iPhoneXR = "iPhone XR",
iPhone11 = "iPhone 11",
iPhone11Pro = "iPhone 11 Pro",
iPhone11ProMax = "iPhone 11 Pro Max",
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,
"iPhone11,2" : .iPhoneXS,
"iPhone11,4" : .iPhoneXSmax,
"iPhone11,6" : .iPhoneXSmax,
"iPhone11,8" : .iPhoneXR,
"iPhone12,1" : .iPhone11,
"iPhone12,3" : .iPhone11Pro,
"iPhone12,5" : .iPhone11ProMax
]
if let model = modelMap[String.init(validatingUTF8: modelCode!)!] {
return model
}
return Model.unrecognized
}
}
在Swift 3中是这样的
UIDevice.current.model
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)
}
当你使用Swift 3时，接受的答案有一些问题! 这个答案(灵感来自NAZIK)适用于Swift 3和新款iPhone:
import UIKit
public extension UIDevice {
var modelName: String {
#if (arch(i386) || arch(x86_64)) && os(iOS)
let DEVICE_IS_SIMULATOR = true
#else
let DEVICE_IS_SIMULATOR = false
#endif
var machineString = String()
if DEVICE_IS_SIMULATOR == true
{
if let dir = ProcessInfo().environment["SIMULATOR_MODEL_IDENTIFIER"] {
machineString = dir
}
}
else {
var systemInfo = utsname()
uname(&systemInfo)
let machineMirror = Mirror(reflecting: systemInfo.machine)
machineString = 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 machineString {
case "iPod4,1": return "iPod Touch 4G"
case "iPod5,1": return "iPod Touch 5G"
case "iPod7,1": return "iPod Touch 6G"
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 "iPhone8,4": return "iPhone SE"
case "iPhone9,1", "iPhone9,3": return "iPhone 7"
case "iPhone9,2", "iPhone 9,4": return "iPhone 7 Plus"
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 "iPad6,3", "iPad6,4": return "iPad Pro (9.7 inch)"
case "iPad6,7", "iPad6,8": return "iPad Pro (12.9 inch)"
case "AppleTV5,3": return "Apple TV"
default: return machineString
}
}
}
获取模型名称(营销名称)的最简单方法
小心使用私有API -[UIDevice _deviceinfokey:]，你不会被Apple拒绝。
// works on both simulators and real devices, iOS 8 to iOS 12
NSString *deviceModelName(void) {
// For Simulator
NSString *modelName = NSProcessInfo.processInfo.environment[@"SIMULATOR_DEVICE_NAME"];
if (modelName.length > 0) {
return modelName;
}
// For real devices and simulators, except simulators running on iOS 8.x
UIDevice *device = [UIDevice currentDevice];
NSString *selName = [NSString stringWithFormat:@"_%@ForKey:", @"deviceInfo"];
SEL selector = NSSelectorFromString(selName);
if ([device respondsToSelector:selector]) {
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Warc-performSelector-leaks"
modelName = [device performSelector:selector withObject:@"marketing-name"];
#pragma clang diagnostic pop
}
return modelName;
}
我是怎么得到"marketing-name"这个密钥的?
运行在模拟器上的NSProcessInfo.processInfo.environment包含一个名为“SIMULATOR_CAPABILITIES”的键，其值是一个plist文件。然后你打开plist文件，你会得到模型名称的关键字“marketing-name”。
