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

这里有一个没有强制unwrap和Swift 3.0的修改:

import Foundation
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",
    iPhone6        = "iPhone 6",
    iPhone6plus    = "iPhone 6 Plus",
    iPhone6S       = "iPhone 6S",
    iPhone6Splus   = "iPhone 6S Plus",
    iPhoneSE       = "iPhone SE",
    iPhone7        = "iPhone 7",
    iPhone7plus    = "iPhone 7 Plus",
    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,
            "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,
            ]

        guard let safeModelCode = modelCode else {
            return Model.unrecognized
        }

        guard let modelString = String.init(validatingUTF8: safeModelCode) else {
            return Model.unrecognized
        }

        guard let model = modelMap[modelString] else {
            return Model.unrecognized
        }

        return model
    }
}

其他回答

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。

extension UIDevice {

    public static let hardwareModel: String = {
        var path = [CTL_HW, HW_MACHINE]
        var n = 0
        sysctl(&path, 2, nil, &n, nil, 0)
        var a: [UInt8] = .init(repeating: 0, count: n)
        sysctl(&path, 2, &a, &n, nil, 0)
        return .init(cString: a)
    }()
}

UIDevice.hardwareModel // → iPhone9,3

这是一个用于检测apple设备的新库

import DeviceDetector

let detector = DeviceDetector.shared
let deviceName = detector.currentDeviceName
let deviceSet = detector.currentDevice
        
let information = """
Model: \(deviceName)
iPhone?: \(detector.isiPhone)
iPad?: \(detector.isiPad)
Notch?: \(detector.hasSafeArea)
        
4inch?: \(DeviceSet.iPhone4inchSet.contains(deviceSet))
4.7inch?: \(DeviceSet.iPhone4_7inchSet.contains(deviceSet))
iPhoneSE?: \(DeviceSet.iPhoneSESet.contains(deviceSet))
iPhonePlus?: \(DeviceSet.iPhonePlusSet.contains(deviceSet))
iPadPro?: \(DeviceSet.iPadProSet.contains(deviceSet))
"""

结果

如果你不想每次苹果向设备家族添加新型号时都更新你的代码,可以使用下面的方法只返回型号代码。

func platform() -> String {
        var systemInfo = utsname()
        uname(&systemInfo)
        let modelCode = withUnsafeMutablePointer(&systemInfo.machine) {
            ptr in String.fromCString(UnsafePointer<CChar>(ptr))
        }

        return String.fromCString(modelCode!)!
}

对我有用

var sysinfo = utsname()
uname(&sysinfo)
let data = Data(bytes: &sysinfo.machine, count: Int(_SYS_NAMELEN))
let model = String(cString: [UInt8](data) //iPhone13,1