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

斯威夫特5

/// Obtain the machine hardware platform from the `uname()` unix command
///
/// Example of return values
///  - `"iPhone8,1"` = iPhone 6s
///  - `"iPad6,7"` = iPad Pro (12.9-inch)
static var unameMachine: String {
    var utsnameInstance = utsname()
    uname(&utsnameInstance)
    let optionalString: String? = withUnsafePointer(to: &utsnameInstance.machine) {
        $0.withMemoryRebound(to: CChar.self, capacity: 1) {
            ptr in String.init(validatingUTF8: ptr)
        }
    }
    return optionalString ?? "N/A"
}

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()

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")
    }
struct utsname systemInfo;
uname(&systemInfo);

NSString* deviceModel = [NSString stringWithCString:systemInfo.machine
                          encoding:NSUTF8StringEncoding];

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!)!
}