ios 在 Swift 中使用 UI_USER_INTERFACE_IDIOM() 检测当前设备
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时间：2020-08-31 00:31:24 来源:igfitidea点击:
Detect current device with UI_USER_INTERFACE_IDIOM() in Swift
提问by Berry Blue
What is the equivalent of UI_USER_INTERFACE_IDIOM()in Swift to detect between iPhone and iPad?
UI_USER_INTERFACE_IDIOM()在 Swift 中检测 iPhone 和 iPad相当于什么？
I get an Use of unresolved identifiererror when compiling in Swift.
Use of unresolved identifier在 Swift 中编译时出现错误。
回答by Cezar
When working with Swift, you can use the enumUIUserInterfaceIdiom, defined as:
使用 Swift 时，您可以使用enumUIUserInterfaceIdiom, 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)
}
So you can use it as:
因此，您可以将其用作：
UIDevice.current.userInterfaceIdiom == .pad
UIDevice.current.userInterfaceIdiom == .phone
UIDevice.current.userInterfaceIdiom == .unspecified
Or with a Switch statement:
或者使用 Switch 语句：
switch UIDevice.current.userInterfaceIdiom {
case .phone:
// It's an iPhone
case .pad:
// It's an iPad (or macOS Catalyst)
case .unspecified:
// Uh, oh! What could it be?
}
UI_USER_INTERFACE_IDIOM()is an Objective-C macro, which is defined as:
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)
Also, note that even when working with Objective-C, the UI_USER_INTERFACE_IDIOM()macro is only required when targeting iOS 3.2 and below. When deploying to iOS 3.2 and up, you can use [UIDevice userInterfaceIdiom]directly.
另外，请注意，即使使用 Objective-C，UI_USER_INTERFACE_IDIOM()也仅在针对 iOS 3.2 及更低版本时才需要该宏。部署到iOS 3.2及以上时，可以[UIDevice userInterfaceIdiom]直接使用。
回答by Beslan Tularov
You should use this GBDeviceInfoframeworkor ...
您应该使用此GBDeviceInfo框架或...
Apple defines this:
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
}
so for the strict definition of the device can be used this code
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
}
how to use
如何使用
if DeviceType.IS_IPHONE_6P_7P {
print("IS_IPHONE_6P_7P")
}
to detect iOS version
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)
}
how to use
如何使用
if Version.iOS8 {
print("iOS8")
}
回答by user3378170
Swift 2.0 & iOS 9 & Xcode 7.1
Swift 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")
}
Swift 3.0 and Swift 4.0
Swift 3.0 和 Swift 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")
}
Use UITraitCollection. The iOS trait environment is exposed though the traitCollectionproperty of the UITraitEnvironment protocol. This protocol is adopted by the following classes:
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：
- UIScreen
- UIWindow
- UIViewController
- UIPresentationController
- UIView
- 用户界面
- 界面窗口
- 视图控制器
- UIPresentationController
- 界面视图
回答by Masterfego
if/else case:
如果/其他情况：
if (UIDevice.currentDevice().userInterfaceIdiom == UIUserInterfaceIdiom.Pad)
{
// Ipad
}
else
{
// Iphone
}
回答by ricardo
I do in that way:
我是这样做的：
UIDevice.current.model
It shows the name of the device.
它显示设备的名称。
To check if is iPad or iPhone:
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")
}
回答by Alessandro Ornano
Swift 2.x:
斯威夫特 2.x：
Adding to Beslav Turalov answer'sthe new entry iPad Pro can easily be find with this line
添加到Beslav Turalov 答案的新条目 iPad Pro 可以通过这条线轻松找到
to detect iPad Pro
检测 iPad Pro
struct DeviceType
{
...
static let IS_IPAD_PRO = UIDevice.currentDevice().userInterfaceIdiom == .Pad && ScreenSize.SCREEN_MAX_LENGTH == 1366.0
}
Swift 3 (TV and car added):
Swift 3（添加电视和汽车）：
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)
}
USAGE:
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") }
回答by iBug
Try adding an extension like this:
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
}
}
}
This is how you will use it:
这是你将如何使用它：
let modelName = UIDevice.currentDevice().modelName
EDITFor simulator, you can try a solution here
编辑对于模拟器，您可以在此处尝试解决方案
回答by Brody Robertson
Swift 4.2 Extension
Swift 4.2 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
}
}
Usage
用法
if UIDevice.isPad {
// Do something
}
回答by Giang
Thank you everybody support :))
谢谢大家支持：））
UIDevice+Extensions.swift
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
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
}
String+Whitespace.swift
字符串+空白.swift
import Foundation
extension String {
public func trimmed() -> String {
return self.trimmingCharacters(in: .whitespacesAndNewlines)
}
}
回答by Nick
Try this for check current device is iPhone or iPad:
试试这个来检查当前设备是 iPhone 还是 iPad：
Swift 5
斯威夫特 5
struct Device {
static let IS_IPAD = UIDevice.current.userInterfaceIdiom == .pad
static let IS_IPHONE = UIDevice.current.userInterfaceIdiom == .phone
}
Use:
用：
if(Device.IS_IPHONE){
// device is iPhone
}if(Device.IS_IPAD){
// device is iPad (or a Mac running under macOS Catalyst)
}else{
// other
}
