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
其他回答
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
}
struct DeviceType {
static let IS_IPHONE_4_OR_LESS = UIDevice.current.userInterfaceIdiom == .phone && Constants.SCREEN_MAX_LENGTH < 568
static let IS_IPHONE_5 = UIDevice.current.userInterfaceIdiom == .phone && Constants.SCREEN_MAX_LENGTH == 568
static let IS_IPHONE_6 = UIDevice.current.userInterfaceIdiom == .phone && Constants.SCREEN_MAX_LENGTH == 667
static let IS_IPHONE_6P = UIDevice.current.userInterfaceIdiom == .phone && Constants.SCREEN_MAX_LENGTH == 736
static let IS_IPAD = UIDevice.current.userInterfaceIdiom == .pad && Constants.SCREEN_MAX_LENGTH == 1024
}
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
}
这个Swift 3.0示例将当前设备模型作为枚举常量返回(以避免与字符串字面值直接比较)。enum的原始值是一个包含人类可读的iOS设备名称的字符串。因为它是Swift，所以认可的设备列表只包括最新的机型，足以支持包含Swift的iOS版本。下面的使用示例利用了这个答案末尾的实现:
switch UIDevice().type {
case .iPhone5:
print("No TouchID sensor")
case .iPhone5S:
fallthrough
case .iPhone6:
fallthrough
case .iPhone6plus:
fallthrough
case .iPad_Pro9_7:
fallthrough
case .iPad_Pro12_9:
fallthrough
case .iPhone7:
fallthrough
case .iPhone7plus:
print("Put your thumb on the " +
UIDevice().type.rawValue + " TouchID sensor")
case .unrecognized:
print("Device model unrecognized");
default:
print(UIDevice().type.rawValue + " not supported by this app");
}
你的应用应该保持最新的新设备发布，以及当苹果为同一设备家族添加新型号时。例如，iPhone3、1 iPhone3、2 iPhone3、4都是“iPhone 4”。避免编写不考虑新型号的代码，这样你的算法就不会意外地配置失败或对新设备做出响应。你可以参考这个维护列表的iOS设备型号#更新你的应用程序在战略时刻。
iOS includes device-independent interfaces to detect hardware capabilities and parameters such as screen size. The generalized interfaces Apple provides are usually the safest, best supported mechanisms to dynamically adapt an app's behavior to different hardware. Nevertheless, the following code can be useful for prototyping, debugging, testing, or any time code needs to target a specific device family. This technique can also be useful to describe the current device by its common/publicly recognized name.
斯威夫特3
// 1. Declare outside class definition (or in its own file).
// 2. UIKit must be included in file where this code is added.
// 3. Extends UIDevice class, thus is available anywhere in app.
//
// Usage example:
//
// if UIDevice().type == .simulator {
// print("You're running on the simulator... boring!")
// } else {
// print("Wow! Running on a \(UIDevice().type.rawValue)")
// }
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",
iPadPro9_7 = "iPad Pro 9.7\"",
iPadPro9_7_cell = "iPad Pro 9.7\" cellular",
iPadPro10_5 = "iPad Pro 10.5\"",
iPadPro10_5_cell = "iPad Pro 10.5\" cellular",
iPadPro12_9 = "iPad Pro 12.9\"",
iPadPro12_9_cell = "iPad Pro 12.9\" cellular",
iPhone6 = "iPhone 6",
iPhone6plus = "iPhone 6 Plus",
iPhone6S = "iPhone 6S",
iPhone6Splus = "iPhone 6S Plus",
iPhoneSE = "iPhone SE",
iPhone7 = "iPhone 7",
iPhone7plus = "iPhone 7 Plus",
iPhone8 = "iPhone 8",
iPhone8plus = "iPhone 8 Plus",
iPhoneX = "iPhone X",
iPhoneXS = "iPhone XS",
iPhoneXSmax = "iPhone XS Max",
iPhoneXR = "iPhone XR",
iPhone11 = "iPhone 11",
iPhone11Pro = "iPhone 11 Pro",
iPhone11ProMax = "iPhone 11 Pro Max",
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,
"iPad6,3" : .iPadPro9_7,
"iPad6,11" : .iPadPro9_7,
"iPad6,4" : .iPadPro9_7_cell,
"iPad6,12" : .iPadPro9_7_cell,
"iPad6,7" : .iPadPro12_9,
"iPad6,8" : .iPadPro12_9_cell,
"iPad7,3" : .iPadPro10_5,
"iPad7,4" : .iPadPro10_5_cell,
"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,
"iPhone10,1" : .iPhone8,
"iPhone10,2" : .iPhone8plus,
"iPhone10,3" : .iPhoneX,
"iPhone10,6" : .iPhoneX,
"iPhone11,2" : .iPhoneXS,
"iPhone11,4" : .iPhoneXSmax,
"iPhone11,6" : .iPhoneXSmax,
"iPhone11,8" : .iPhoneXR,
"iPhone12,1" : .iPhone11,
"iPhone12,3" : .iPhone11Pro,
"iPhone12,5" : .iPhone11ProMax
]
if let model = modelMap[String.init(validatingUTF8: modelCode!)!] {
return model
}
return Model.unrecognized
}
}
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()
