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")
    }

其他回答

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"

下面是获取硬件字符串的代码,但是您需要比较这些硬件字符串才能知道它是哪个设备。我已经创建了一个类,包含几乎所有的设备字符串(我们保持字符串最新的新设备)。使用方便,请检查

吉hub /恶魔大师

Objective-C: GitHub/DeviceUtil

public func hardwareString() -> String {
  var name: [Int32] = [CTL_HW, HW_MACHINE]
  var size: Int = 2
  sysctl(&name, 2, nil, &size, &name, 0)
  var hw_machine = [CChar](count: Int(size), repeatedValue: 0)
  sysctl(&name, 2, &hw_machine, &size, &name, 0)

  let hardware: String = String.fromCString(hw_machine)!
  return hardware
}

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()

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)