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在 Swift 中使用 UI_USER_INTERFACE_IDIOM() 检测当前设备

在 Swift 中检测 iPhone 和 iPad 之间的 UI_USER_INTERFACE_IDIOM() 的等价物是什么？

在 Swift 中编译时出现 Use of unresolved identifier 错误。

E
EJZ

使用 Swift 时，您可以使用 enum UIUserInterfaceIdiom，定义为： 

enum UIUserInterfaceIdiom : Int {
    case unspecified
    
    case phone // iPhone and iPod touch style UI
    case pad   // iPad style UI (also includes macOS Catalyst)
}

因此，您可以将其用作： 

UIDevice.current.userInterfaceIdiom == .pad
UIDevice.current.userInterfaceIdiom == .phone
UIDevice.current.userInterfaceIdiom == .unspecified

或使用 Switch 语句： 

    switch UIDevice.current.userInterfaceIdiom {
    case .phone:
        // It's an iPhone
    case .pad:
        // It's an iPad (or macOS Catalyst)

     @unknown default:
        // Uh, oh! What could it be?
    }

UI_USER_INTERFACE_IDIOM() 是一个 Objective-C 宏，定义为： 

#define UI_USER_INTERFACE_IDIOM() \ ([[UIDevice currentDevice] respondsToSelector:@selector(userInterfaceIdiom)] ? \ [[UIDevice currentDevice] userInterfaceIdiom] : \ UIUserInterfaceIdiomPhone)

此外，请注意，即使使用 Objective-C，也只有在面向 iOS 3.2 及更低版本时才需要 UI_USER_INTERFACE_IDIOM() 宏。部署到 iOS 3.2 及更高版本时，您可以直接使用 [UIDevice userInterfaceIdiom]。

没关系。我得到了它与 if UIDevice.currentDevice().userInterfaceIdiom == .Pad 的合作
正如托尼在下面的答案之一中提到的那样，当通过 TestFlight 部署应用程序时，Swift 应用程序中的 UI_USER_INTERFACE_IDIOM 会崩溃。奇怪的是，当应用程序从 X-Code 直接上传到设备时，它就可以工作。我也遇到了这个错误。
@Zmey 是的，我的应用也被拒绝了，因为 UI_USER_INTERFACE_IDIOM 在审核中崩溃，很奇怪
在 Swift 3 中，UIDevice.currentDevice().userInterfaceIdiom 变为 UIDevice.current.userInterfaceIdiom
如果您的应用程序仅适用于 iPhone，则此方法将无法正常工作，您将始终获得 .phone，请查看 Ricardo 的答案。
C
Cœur

您应该使用这个 GBDeviceInfo 框架 或...

苹果定义了这一点： 

public enum UIUserInterfaceIdiom : Int {

    case unspecified

    case phone // iPhone and iPod touch style UI

    case pad // iPad style UI

    @available(iOS 9.0, *)
    case tv // Apple TV style UI

    @available(iOS 9.0, *)
    case carPlay // CarPlay style UI
}

所以对于设备的严格定义可以使用这个代码 

struct ScreenSize
{
    static let SCREEN_WIDTH         = UIScreen.main.bounds.size.width
    static let SCREEN_HEIGHT        = UIScreen.main.bounds.size.height
    static let SCREEN_MAX_LENGTH    = max(ScreenSize.SCREEN_WIDTH, ScreenSize.SCREEN_HEIGHT)
    static let SCREEN_MIN_LENGTH    = min(ScreenSize.SCREEN_WIDTH, ScreenSize.SCREEN_HEIGHT)
}

struct DeviceType
{
    static let IS_IPHONE_4_OR_LESS  = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH < 568.0
    static let IS_IPHONE_5          = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 568.0
    static let IS_IPHONE_6_7          = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 667.0
    static let IS_IPHONE_6P_7P         = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 736.0
    static let IS_IPAD              = UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.SCREEN_MAX_LENGTH == 1024.0
    static let IS_IPAD_PRO          = UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.SCREEN_MAX_LENGTH == 1366.0
}

如何使用 

if DeviceType.IS_IPHONE_6P_7P {
    print("IS_IPHONE_6P_7P")
}

检测iOS版本 

struct Version{
    static let SYS_VERSION_FLOAT = (UIDevice.current.systemVersion as NSString).floatValue
    static let iOS7 = (Version.SYS_VERSION_FLOAT < 8.0 && Version.SYS_VERSION_FLOAT >= 7.0)
    static let iOS8 = (Version.SYS_VERSION_FLOAT >= 8.0 && Version.SYS_VERSION_FLOAT < 9.0)
    static let iOS9 = (Version.SYS_VERSION_FLOAT >= 9.0 && Version.SYS_VERSION_FLOAT < 10.0)
}

如何使用 

if Version.iOS8 {
    print("iOS8")
}
我喜欢 struct ScreenSize/DeviceType 方法，因为它适用于模拟器
批准的答案应该去这个整洁的答案
来自印度的厚爱，感谢您的努力，非常感谢您分享和制作更好的 Stackoverflow ;)
更新的代码是什么？在 iPhone 7 和 7P 的 DEVICE_TYPE 上下文中
N
Nikolay Suvandzhiev

如果/其他情况： 

 if UIDevice.current.userInterfaceIdiom == .pad {
     // iPad
 } else {
     // not iPad (iPhone, mac, tv, carPlay, unspecified)
 }
public enum UIUserInterfaceIdiom : Int { case Unspecified @available(iOS 3.2, *) case Phone // iPhone 和 iPod touch 风格 UI @available(iOS 3.2, *) case Pad // iPad 风格 UI @available(iOS 9.0, *) case TV // Apple TV 风格的 UI } 查看 UIUserInterfaceIdiom 的定义。如果不是 Pad，可能是 Phone、TV、Unspecified。
J
Jonas Deichelmann

斯威夫特 2.0 & iOS 9 & Xcode 7.1 

// 1. request an UITraitCollection instance
let deviceIdiom = UIScreen.mainScreen().traitCollection.userInterfaceIdiom

// 2. check the idiom
switch (deviceIdiom) {

case .Pad:
    print("iPad style UI")
case .Phone:
    print("iPhone and iPod touch style UI")
case .TV: 
    print("tvOS style UI")
default:
    print("Unspecified UI idiom")

}

斯威夫特 3.0 和斯威夫特 4.0 

// 1. request an UITraitCollection instance
let deviceIdiom = UIScreen.main.traitCollection.userInterfaceIdiom

// 2. check the idiom
switch (deviceIdiom) {

case .pad:
    print("iPad style UI")
case .phone:
    print("iPhone and iPod touch style UI")
case .tv: 
    print("tvOS style UI")
default:
    print("Unspecified UI idiom")
}

使用 UITraitCollection。 iOS trait 环境通过 UITraitEnvironment 协议的 traitCollection 属性公开。该协议被以下类采用：

用户界面

界面窗口

UIViewController

UIPresentationController

界面视图

r
ricardo

我这样做： 

UIDevice.current.model

它显示设备的名称。

要检查是 iPad 还是 iPhone： 

if ( UIDevice.current.model.range(of: "iPad") != nil){
    print("I AM IPAD")
} else {
    print("I AM IPHONE")
}
绝对是最好的解决方案，至少对我来说。检查 userInterfaceIdiom 是否存在问题：如果您的应用仅适用于 iPhone，但您在 iPad 上启动应用，则 userInterfaceIdiom 为 == .Phone
绝对是最好的解决方案。好一个。
B
Brody Robertson

Swift 4.2 - 5.1 扩展 

 public extension UIDevice {

    class var isPhone: Bool {
        return UIDevice.current.userInterfaceIdiom == .phone
    }

    class var isPad: Bool {
        return UIDevice.current.userInterfaceIdiom == .pad
    }

    class var isTV: Bool {
        return UIDevice.current.userInterfaceIdiom == .tv
    }

    class var isCarPlay: Bool {
        return UIDevice.current.userInterfaceIdiom == .carPlay
    }
}

用法 

if UIDevice.isPad {
   // Do something
}
C
Community

尝试添加这样的扩展： 

    public extension UIDevice {

    var 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 where value != 0 else { return identifier }
            return identifier + String(UnicodeScalar(UInt8(value)))
        }

        switch identifier {
        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 "iPhone9,1", "iPhone9,3":                  return "iPhone 7"
        case "iPhone9,2", "iPhone9,4":                  return "iPhone 7 Plus"
        case "iPhone8,4":                               return "iPhone SE"
        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 "iPad6,3", "iPad6,4", "iPad6,7", "iPad6,8":return "iPad Pro"
        case "AppleTV5,3":                              return "Apple TV"
        case "i386", "x86_64":                          return "Simulator"
        default:                                        return identifier
        }
    }

}

这是你将如何使用它： 

let modelName = UIDevice.currentDevice().modelName

编辑 对于模拟器，您可以尝试解决方案here

很好，虽然用模拟器测试只会返回模拟器。有没有解决的办法？
C
Community

斯威夫特 2.x：

添加到 Beslav Turalov answer's 的新条目 iPad Pro 可以通过此行轻松找到

检测 iPad Pro 

struct DeviceType
{
    ...
    static let IS_IPAD_PRO = UIDevice.currentDevice().userInterfaceIdiom == .Pad && ScreenSize.SCREEN_MAX_LENGTH == 1366.0
}

Swift 3（添加了电视和汽车）： 

struct ScreenSize
{
    static let SCREEN_WIDTH         = UIScreen.main.bounds.size.width
    static let SCREEN_HEIGHT        = UIScreen.main.bounds.size.height
    static let SCREEN_MAX_LENGTH    = max(ScreenSize.SCREEN_WIDTH, ScreenSize.SCREEN_HEIGHT)
    static let SCREEN_MIN_LENGTH    = min(ScreenSize.SCREEN_WIDTH, ScreenSize.SCREEN_HEIGHT)
}

struct DeviceType
{
    static let IS_IPHONE            = UIDevice.current.userInterfaceIdiom == .phone
    static let IS_IPHONE_4_OR_LESS  = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH < 568.0
    static let IS_IPHONE_5          = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 568.0
    static let IS_IPHONE_6          = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 667.0
    static let IS_IPHONE_6P         = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 736.0
    static let IS_IPHONE_7          = IS_IPHONE_6
    static let IS_IPHONE_7P         = IS_IPHONE_6P
    static let IS_IPAD              = UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.SCREEN_MAX_LENGTH == 1024.0
    static let IS_IPAD_PRO_9_7      = IS_IPAD
    static let IS_IPAD_PRO_12_9     = UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.SCREEN_MAX_LENGTH == 1366.0
    static let IS_TV                = UIDevice.current.userInterfaceIdiom == .tv
    static let IS_CAR_PLAY          = UIDevice.current.userInterfaceIdiom == .carPlay
}

struct Version{
    static let SYS_VERSION_FLOAT = (UIDevice.current.systemVersion as NSString).floatValue
    static let iOS7 = (Version.SYS_VERSION_FLOAT < 8.0 && Version.SYS_VERSION_FLOAT >= 7.0)
    static let iOS8 = (Version.SYS_VERSION_FLOAT >= 8.0 && Version.SYS_VERSION_FLOAT < 9.0)
    static let iOS9 = (Version.SYS_VERSION_FLOAT >= 9.0 && Version.SYS_VERSION_FLOAT < 10.0)
    static let iOS10 = (Version.SYS_VERSION_FLOAT >= 10.0 && Version.SYS_VERSION_FLOAT < 11.0)
}

用法： 

if DeviceType.IS_IPHONE_7P { print("iPhone 7 plus") }
if DeviceType.IS_IPAD_PRO_9_7 && Version.iOS10 { print("iPad pro 9.7 with iOS 10 version") }
r
raaz

在 swift 4 & Xcode 9.2 中，您可以通过以下方式检测设备是否为 iPhone/iPad。 

if (UIDevice.current.userInterfaceIdiom == .pad){
   print("iPad")
}
else{
   print("iPhone")
}

另一种方式 

    let deviceName = UIDevice.current.model
    print(deviceName);
    if deviceName == "iPhone"{
        print("iPhone")
    }
    else{
        print("iPad")
    }
d
drewster

试试这个检查当前设备是 iPhone 还是 iPad：

斯威夫特 5 

struct Device {
    static let IS_IPAD = UIDevice.current.userInterfaceIdiom == .pad
    static let IS_IPHONE = UIDevice.current.userInterfaceIdiom == .phone
}

利用： 

if(Device.IS_IPHONE){
    // device is iPhone
}if(Device.IS_IPAD){
    // device is iPad (or a Mac running under macOS Catalyst)
}else{
    // other
}
A
Abdullah Umer

自 iOS 13 起，UI_USER_INTERFACE_IDIOM 已被弃用。如果您的代码仍在 Obj-C 中，您可以使用以下代码： 

if (UIDevice.currentDevice.userInterfaceIdiom == UIUserInterfaceIdiomPad) {
    // device is iPad
}

在哪里： 

typedef NS_ENUM(NSInteger, UIUserInterfaceIdiom) {
    UIUserInterfaceIdiomUnspecified = -1,
    UIUserInterfaceIdiomPhone API_AVAILABLE(ios(3.2)), // iPhone and iPod touch style UI
    UIUserInterfaceIdiomPad API_AVAILABLE(ios(3.2)), // iPad style UI
    UIUserInterfaceIdiomTV API_AVAILABLE(ios(9.0)), // Apple TV style UI
    UIUserInterfaceIdiomCarPlay API_AVAILABLE(ios(9.0)), // CarPlay style UI
};
感谢您的 Obj-C 回答。
更多 Obj-C 对齐写法 - [[UIDevice currentDevice] userInterfaceIdiom]
G
Giang

谢谢大家的支持：））

UIDevice+Extensions.swift 

import Foundation
import UIKit

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 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 "iPhone9,1", "iPhone9,3":                  return "iPhone 7"
            case "iPhone9,2", "iPhone9,4":                  return "iPhone 7 Plus"
            case "iPhone8,4":                               return "iPhone SE"
            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 "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 "iPad6,11", "iPad6,12":                    return "iPad 5"
            case "iPad7,5", "iPad7,6":                      return "iPad 6"
            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 "iPad6,3", "iPad6,4":                      return "iPad Pro 9.7 Inch"
            case "iPad6,7", "iPad6,8":                      return "iPad Pro 12.9 Inch"
            case "iPad7,1", "iPad7,2":                      return "iPad Pro 12.9 Inch 2. Generation"
            case "iPad7,3", "iPad7,4":                      return "iPad Pro 10.5 Inch"
            case "AppleTV5,3":                              return "Apple TV"
            case "AppleTV6,2":                              return "Apple TV 4K"
            case "AudioAccessory1,1":                       return "HomePod"
            case "i386", "x86_64":                          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)
    }()
}

enum DeviceName: String {
    case iPod_Touch_5 = "iPod Touch 5"
    case pod_Touch_6 = "Pod Touch 6"
    case iPhone_4 = "iPhone 4"
    case iPhone_4s = "iPhone 4s"
    case iPhone_5 = "iPhone 5"
    case iPhone_5c = "iPhone 5c"
    case iPhone_5s = "iPhone 5s"
    case iPhone_6 = "iPhone 6"
    case iPhone_6_Plus = "iPhone 6 Plus"
    case iPhone_6s = "iPhone 6s"
    case iPhone_6s_Plus = "iPhone 6s Plus"
    case iPhone_7 = "iPhone 7"
    case iPhone_7_Plus = "iPhone 7 Plus"
    case iPhone_SE = "iPhone SE"
    case iPhone_8 = "iPhone 8"
    case iPhone_8_Plus = "iPhone 8 Plus"
    case iPhone_X = "iPhone X"
    case iPhone_XS = "iPhone XS"
    case iPhone_XS_Max = "iPhone XS Max"
    case iPhone_XR = "iPhone XR"
    case iPad_2 = "iPad 2"
    case iPad_3 = "iPad 3"
    case iPad_4 = "iPad 4"
    case iPad_Air = "iPad Air"
    case iPad_Air_2 = "iPad Air 2"
    case iPad_5 = "iPad 5"
    case iPad_6 = "iPad 6"
    case iPad_Mini = "iPad Mini"
    case iPad_Mini_2 = "iPad Mini 2"
    case iPad_Mini_3 = "iPad Mini 3"
    case iPad_Mini_4 = "iPad Mini 4"
    case iPad_Pro_9_7_Inch = "iPad Pro 9.7 Inch"
    case iPad_Pro_12_9_Inch = "iPad Pro 12.9 Inch"
    case iPad_Pro_12_9_Inch_2_Generation = "iPad Pro 12.9 Inch 2. Generation"
    case iPad_Pro_10_5_Inch = "iPad Pro 10.5 Inch"
    case apple_TV = "Apple TV"
    case apple_TV_4K = "Apple TV 4K"
    case homePod = "HomePod"
}

SharedFunctions.swift 

import Foundation
import UIKit
func isDevice(_ name: DeviceName) -> Bool {
    let modelName = UIDevice.modelName.replacingOccurrences(of: "Simulator", with: "").trimmed()
    if name.rawValue == modelName {
        return true
    }

    return false
}

字符串+空格.swift 

import Foundation

extension String {
   public func trimmed() -> String {
    return self.trimmingCharacters(in: .whitespacesAndNewlines)
  }
}
Y
YannSteph

Swift 2.0 & iOS 7+ / iOS 8+ / iOS 9+ 

public class Helper {
    public class var isIpad:Bool {
        if #available(iOS 8.0, *) {
            return UIScreen.mainScreen().traitCollection.userInterfaceIdiom == .Pad
        } else {
            return UIDevice.currentDevice().userInterfaceIdiom == .Pad
        }
    }
    public class var isIphone:Bool {
        if #available(iOS 8.0, *) {
            return UIScreen.mainScreen().traitCollection.userInterfaceIdiom == .Phone
        } else {
            return UIDevice.currentDevice().userInterfaceIdiom == .Phone
        }
    }
}

利用 ： 

if Helper.isIpad {

}

或者 

guard Helper.isIpad else {
    return
}

谢谢@user3378170

感谢@user3378170 支持 iOS 9
K
Kevin

仅供参考，我已将 UI_USER_INTERFACE_IDIOM() 用于我用 Swift 编写的应用程序。该应用程序可以使用 XCode 6.3.1 很好地编译，而不会对该命令发出任何警告，在模拟器（使用任何选定的设备）和我所有的 iOS 版本从 7.1 到 8.3 的真实设备（iPhone、iPad）上运行良好。

但是，该应用程序在 Apple 审阅者的设备上崩溃（并被拒绝）。我花了几天的时间才发现问题，因为很少重新上传到 iTunes Connect。

现在我改用 UIDevice.currentDevice().userInterfaceIdiom，我的应用程序可以从此类崩溃中幸存下来。

完全正确。它确实崩溃了，让我很头疼，试图找出问题所在。
每当我在代码中使用 UI_USER_INTERFACE_IDIOM() 时，Swift 编译器就会一直崩溃，而没有任何错误消息。很奇怪。
fwiw，Apple 的文档现在声明“如果您的应用程序在 iOS 3.2 及更高版本中运行，请改用 userInterfaceIdiom。”
s
selva

如果您想检查当前设备是 iPad 还是 iPhone，那么您可以使用以下代码行： 

 if(UIDevice.currentDevice().userInterfaceIdiom == .Pad){

  }else if(UIDevice.currentDevice().userInterfaceIdiom == .Phone){

  }
o
odemolliens

斯威夫特 3.0： 

let userInterface = UIDevice.current.userInterfaceIdiom

if(userInterface == .pad){
    //iPads
}else if(userInterface == .phone){
    //iPhone
}else if(userInterface == .carPlay){
    //CarPlay
}else if(userInterface == .tv){
    //AppleTV
}
a
aBikis

对上述答案进行了一些补充，以便您返回一个类型而不是字符串值。

我认为这主要用于 UI 调整，所以我认为包含所有子模型（即 iPhone 5s）并不相关，但这可以通过在 isDevice 数组中添加模型测试来轻松扩展

测试在 Swift 3.1 Xcode 8.3.2 中使用物理和模拟器设备

执行：

UIDevice.whichDevice() 

public enum SVNDevice {
  case isiPhone4, isIphone5, isIphone6or7, isIphone6por7p, isIphone, isIpad, isIpadPro
}

extension UIDevice {
  class func whichDevice() -> SVNDevice? {
    let isDevice = { (comparision: Array<(Bool, SVNDevice)>) -> SVNDevice? in
      var device: SVNDevice?
      comparision.forEach({
        device = $0.0 ? $0.1 : device
      })
      return device
    }

    return isDevice([
      (UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH < 568.0, SVNDevice.isiPhone4),
      (UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 568.0, SVNDevice.isIphone5),
      (UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 667.0, SVNDevice.isIphone6or7),
      (UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 736.0, SVNDevice.isIphone6por7p),
      (UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.SCREEN_MAX_LENGTH == 1024.0, SVNDevice.isIpad),
      (UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.SCREEN_MAX_LENGTH == 1366.0, SVNDevice.isIpadPro)])
  }
}



private struct ScreenSize {
  static let SCREEN_WIDTH         = UIScreen.main.bounds.size.width
  static let SCREEN_HEIGHT        = UIScreen.main.bounds.size.height
  static let SCREEN_MAX_LENGTH    = max(ScreenSize.SCREEN_WIDTH, ScreenSize.SCREEN_HEIGHT)
  static let SCREEN_MIN_LENGTH    = min(ScreenSize.SCREEN_WIDTH, ScreenSize.SCREEN_HEIGHT)
}

我创建了一个名为 SVNBootstaper 的框架，其中包括此协议和其他一些辅助协议，它是公开的，可通过 Carthage 获得。

A
Andre Goncalves

您可以在 Swift 5 上使用新方法： 

switch traitCollection.userInterfaceIdiom {
        
    case .unspecified:
        // do something
    case .phone:
        // do something
    case .pad:
        // do something
    case .tv:
        // do something
    case .carPlay:
        // do something
    case .mac:
        // do something
    @unknown default:
        // do something
}
这在 Swift 5 中并不是新的，并且已经发布，例如 here。

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