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
Swift 3.0或更高版本
import UIKit
class ViewController: UIViewController {
let device = UIDevice.current
override func viewDidLoad() {
super.viewDidLoad()
let model = device.model
print(model) // e.g. "iPhone"
let modelName = device.modelName
print(modelName) // e.g. "iPhone 6" /* see the extension */
let deviceName = device.name
print(deviceName) // e.g. "My iPhone"
let systemName = device.systemName
print(systemName) // e.g. "iOS"
let systemVersion = device.systemVersion
print(systemVersion) // e.g. "10.3.2"
if let identifierForVendor = device.identifierForVendor {
print(identifierForVendor) // e.g. "E1X2XX34-5X6X-7890-123X-XXX456C78901"
}
}
}
并添加以下扩展
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 "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 "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 "i386", "x86_64": return "Simulator"
default: return identifier
}
}
}
其他回答
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)
}
斯威夫特3.1
我对简单地调用utsname的看法是:
func platform() -> String {
var systemInfo = utsname()
uname(&systemInfo)
let size = Int(_SYS_NAMELEN) // is 32, but posix AND its init is 256....
let s = withUnsafeMutablePointer(to: &systemInfo.machine) {p in
p.withMemoryRebound(to: CChar.self, capacity: size, {p2 in
return String(cString: p2)
})
}
return s
}
和其他的一样，但对C/Swift和后面的复杂结构更清晰一些。 )：
返回值如"x86_64"
我已经实现了一个超轻量级的库，可以根据给出的一些答案来检测使用的设备:https://github.com/schickling/Device.swift
它可以通过迦太基安装，并像这样使用:
import Device
let deviceType = UIDevice.currentDevice().deviceType
switch deviceType {
case .IPhone6: print("Do stuff for iPhone6")
case .IPadMini: print("Do stuff for iPad mini")
default: print("Check other available cases of DeviceType")
}
对于设备和模拟器， 创建一个名为UIDevice.swift的新swift文件
添加以下代码
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 = NSProcessInfo().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 where value != 0 else { return identifier }
return identifier + String(UnicodeScalar(UInt8(value)))
}
}
switch machineString {
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 "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"
default: return machineString
}
}
}
然后在你的视图控制器中，
let deviceType = UIDevice.currentDevice().modelName
if deviceType.lowercaseString.rangeOfString("iphone 4") != nil {
print("iPhone 4 or iphone 4s")
}
else if deviceType.lowercaseString.rangeOfString("iphone 5") != nil {
print("iPhone 5 or iphone 5s or iphone 5c")
}
else if deviceType.lowercaseString.rangeOfString("iphone 6") != nil {
print("iPhone 6 Series")
}
在swift中处理c结构体是很痛苦的。尤其是当里面有c数组的时候。以下是我的解决方案:继续使用objective-c。只需创建一个包装器objective-c类来完成这项工作，然后在swift中使用该类。下面是一个示例类，它就是这样做的:
@interface DeviceInfo : NSObject
+ (NSString *)model;
@end
#import "DeviceInfo.h"
#import <sys/utsname.h>
@implementation DeviceInfo
+ (NSString *)model
{
struct utsname systemInfo;
uname(&systemInfo);
return [NSString stringWithCString: systemInfo.machine encoding: NSUTF8StringEncoding];
}
@end
在迅捷的一面:
let deviceModel = DeviceInfo.model()
