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

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)

其他回答

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()

在Swift 3中是这样的

 UIDevice.current.model

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

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!)!
}

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
    }
}