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

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
    }

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"

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

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()

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"
}