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

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
    }
}
}

其他回答

对于swift4.0及以上使用以下代码:

let udid = UIDevice.current.identifierForVendor?.uuidString
let name = UIDevice.current.name
let version = UIDevice.current.systemVersion
let modelName = UIDevice.current.model
let osName = UIDevice.current.systemName
let localized = UIDevice.current.localizedModel

print(udid ?? "")
print(name)
print(version)
print(modelName)
print(osName)
print(localized)

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()

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

这里有一个辅助库。

斯威夫特5

吊舱“股息”,“~> 2.0”

Swift 4.0 - Swift 4.2

吊舱“股息”,“~> 1.3”

如果你只是想确定模型并做出相应的东西。

你可以这样用:

let isIphoneX = Device().isOneOf([.iPhoneX, .simulator(.iPhoneX)])

在函数中:

func isItIPhoneX() -> Bool {
    let device = Device()
    let check = device.isOneOf([.iPhoneX, .iPhoneXr , .iPhoneXs , .iPhoneXsMax ,
                                .simulator(.iPhoneX), .simulator(.iPhoneXr) , .simulator(.iPhoneXs) , .simulator(.iPhoneXsMax) ])
    return check
}

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。