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
获取模型名称(营销名称)的最简单方法
小心使用私有API -[UIDevice _deviceinfokey:]，你不会被Apple拒绝。
// works on both simulators and real devices, iOS 8 to iOS 12
NSString *deviceModelName(void) {
// For Simulator
NSString *modelName = NSProcessInfo.processInfo.environment[@"SIMULATOR_DEVICE_NAME"];
if (modelName.length > 0) {
return modelName;
}
// For real devices and simulators, except simulators running on iOS 8.x
UIDevice *device = [UIDevice currentDevice];
NSString *selName = [NSString stringWithFormat:@"_%@ForKey:", @"deviceInfo"];
SEL selector = NSSelectorFromString(selName);
if ([device respondsToSelector:selector]) {
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Warc-performSelector-leaks"
modelName = [device performSelector:selector withObject:@"marketing-name"];
#pragma clang diagnostic pop
}
return modelName;
}
我是怎么得到"marketing-name"这个密钥的?
运行在模拟器上的NSProcessInfo.processInfo.environment包含一个名为“SIMULATOR_CAPABILITIES”的键，其值是一个plist文件。然后你打开plist文件，你会得到模型名称的关键字“marketing-name”。
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()
我在UIDevice上做了这个“纯Swift”扩展。
如果你正在寻找一个更优雅的解决方案，你可以使用我在GitHub上发布的µ-框架DeviceKit(也可以通过CocoaPods, Carthage和Swift Package Manager获得)。
代码如下:
import UIKit
public extension UIDevice {
static let modelName: String = {
var systemInfo = utsname()
uname(&systemInfo)
let machineMirror = Mirror(reflecting: systemInfo.machine)
let identifier = machineMirror.children.reduce("") { identifier, element in
guard let value = element.value as? Int8, value != 0 else { return identifier }
return identifier + String(UnicodeScalar(UInt8(value)))
}
func mapToDevice(identifier: String) -> String { // swiftlint:disable:this cyclomatic_complexity
#if os(iOS)
switch identifier {
case "iPod5,1": return "iPod touch (5th generation)"
case "iPod7,1": return "iPod touch (6th generation)"
case "iPod9,1": return "iPod touch (7th generation)"
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 "iPhone9,1", "iPhone9,3": return "iPhone 7"
case "iPhone9,2", "iPhone9,4": return "iPhone 7 Plus"
case "iPhone10,1", "iPhone10,4": return "iPhone 8"
case "iPhone10,2", "iPhone10,5": return "iPhone 8 Plus"
case "iPhone10,3", "iPhone10,6": return "iPhone X"
case "iPhone11,2": return "iPhone XS"
case "iPhone11,4", "iPhone11,6": return "iPhone XS Max"
case "iPhone11,8": return "iPhone XR"
case "iPhone12,1": return "iPhone 11"
case "iPhone12,3": return "iPhone 11 Pro"
case "iPhone12,5": return "iPhone 11 Pro Max"
case "iPhone13,1": return "iPhone 12 mini"
case "iPhone13,2": return "iPhone 12"
case "iPhone13,3": return "iPhone 12 Pro"
case "iPhone13,4": return "iPhone 12 Pro Max"
case "iPhone14,4": return "iPhone 13 mini"
case "iPhone14,5": return "iPhone 13"
case "iPhone14,2": return "iPhone 13 Pro"
case "iPhone14,3": return "iPhone 13 Pro Max"
case "iPhone14,7": return "iPhone 14"
case "iPhone14,8": return "iPhone 14 Plus"
case "iPhone15,2": return "iPhone 14 Pro"
case "iPhone15,3": return "iPhone 14 Pro Max"
case "iPhone8,4": return "iPhone SE"
case "iPhone12,8": return "iPhone SE (2nd generation)"
case "iPhone14,6": return "iPhone SE (3rd generation)"
case "iPad2,1", "iPad2,2", "iPad2,3", "iPad2,4": return "iPad 2"
case "iPad3,1", "iPad3,2", "iPad3,3": return "iPad (3rd generation)"
case "iPad3,4", "iPad3,5", "iPad3,6": return "iPad (4th generation)"
case "iPad6,11", "iPad6,12": return "iPad (5th generation)"
case "iPad7,5", "iPad7,6": return "iPad (6th generation)"
case "iPad7,11", "iPad7,12": return "iPad (7th generation)"
case "iPad11,6", "iPad11,7": return "iPad (8th generation)"
case "iPad12,1", "iPad12,2": return "iPad (9th generation)"
case "iPad13,18", "iPad13,19": return "iPad (10th generation)"
case "iPad4,1", "iPad4,2", "iPad4,3": return "iPad Air"
case "iPad5,3", "iPad5,4": return "iPad Air 2"
case "iPad11,3", "iPad11,4": return "iPad Air (3rd generation)"
case "iPad13,1", "iPad13,2": return "iPad Air (4th generation)"
case "iPad13,16", "iPad13,17": return "iPad Air (5th generation)"
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 "iPad11,1", "iPad11,2": return "iPad mini (5th generation)"
case "iPad14,1", "iPad14,2": return "iPad mini (6th generation)"
case "iPad6,3", "iPad6,4": return "iPad Pro (9.7-inch)"
case "iPad7,3", "iPad7,4": return "iPad Pro (10.5-inch)"
case "iPad8,1", "iPad8,2", "iPad8,3", "iPad8,4": return "iPad Pro (11-inch) (1st generation)"
case "iPad8,9", "iPad8,10": return "iPad Pro (11-inch) (2nd generation)"
case "iPad13,4", "iPad13,5", "iPad13,6", "iPad13,7": return "iPad Pro (11-inch) (3rd generation)"
case "iPad14,3", "iPad14,4": return "iPad Pro (11-inch) (4th generation)"
case "iPad6,7", "iPad6,8": return "iPad Pro (12.9-inch) (1st generation)"
case "iPad7,1", "iPad7,2": return "iPad Pro (12.9-inch) (2nd generation)"
case "iPad8,5", "iPad8,6", "iPad8,7", "iPad8,8": return "iPad Pro (12.9-inch) (3rd generation)"
case "iPad8,11", "iPad8,12": return "iPad Pro (12.9-inch) (4th generation)"
case "iPad13,8", "iPad13,9", "iPad13,10", "iPad13,11":return "iPad Pro (12.9-inch) (5th generation)"
case "iPad14,5", "iPad14,6": return "iPad Pro (12.9-inch) (6th generation)"
case "AppleTV5,3": return "Apple TV"
case "AppleTV6,2": return "Apple TV 4K"
case "AudioAccessory1,1": return "HomePod"
case "AudioAccessory5,1": return "HomePod mini"
case "i386", "x86_64", "arm64": return "Simulator \(mapToDevice(identifier: ProcessInfo().environment["SIMULATOR_MODEL_IDENTIFIER"] ?? "iOS"))"
default: return identifier
}
#elseif os(tvOS)
switch identifier {
case "AppleTV5,3": return "Apple TV 4"
case "AppleTV6,2": return "Apple TV 4K"
case "i386", "x86_64": return "Simulator \(mapToDevice(identifier: ProcessInfo().environment["SIMULATOR_MODEL_IDENTIFIER"] ?? "tvOS"))"
default: return identifier
}
#endif
}
return mapToDevice(identifier: identifier)
}()
}
你这样称呼它:
let modelName = UIDevice.modelName
对于真实的设备，它返回例如:“iPad Pro(12.9英寸)(第五代)”，对于模拟器，它返回例如。模拟式iPad Pro(12.9英寸)(第五代)
下面是模型参考:
https://theiphonewiki.com/wiki/Models https://theiphonewiki.com/wiki/BORD
我已经实现了一个超轻量级的库，可以根据给出的一些答案来检测使用的设备:https://github.com/schickling/Device.swift
它可以通过迦太基安装，并像这样使用:
import Device
let deviceType = UIDevice.currentDevice().deviceType
switch deviceType {
case .IPhone6: print("Do stuff for iPhone6")
case .IPadMini: print("Do stuff for iPad mini")
default: print("Check other available cases of DeviceType")
}
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。
