什么是等价的UI_USER_INTERFACE_IDIOM()在Swift检测之间的iPhone和iPad?
我得到一个使用未解决的标识符错误时，在Swift编译。
当前回答
当使用Swift时，你可以使用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)
}
所以你可以这样使用它:
UIDevice.current.userInterfaceIdiom == .pad
UIDevice.current.userInterfaceIdiom == .phone
UIDevice.current.userInterfaceIdiom == .unspecified
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)
@unknown default:
// Uh, oh! What could it be?
}
UI_USER_INTERFACE_IDIOM()是一个Objective-C宏，它被定义为:
#define UI_USER_INTERFACE_IDIOM() \ ([[UIDevice currentDevice] respondsToSelector:@selector(userInterfaceIdiom)] ? \ [[UIDevice currentDevice] userInterfaceIdiom] : \ UIUserInterfaceIdiomPhone)
还要注意，即使在使用Objective-C时，UI_USER_INTERFACE_IDIOM()宏也只在针对iOS 3.2及以下时才需要。当部署到iOS 3.2及以上版本时，可以直接使用[UIDevice userInterfaceIdiom]。
其他回答
谢谢大家的支持:)
斯威夫特Extensions UIDevice +。
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
}
swift管柱+ Whitespace。
import Foundation
extension String {
public func trimmed() -> String {
return self.trimmingCharacters(in: .whitespacesAndNewlines)
}
}
当使用Swift时，你可以使用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)
}
所以你可以这样使用它:
UIDevice.current.userInterfaceIdiom == .pad
UIDevice.current.userInterfaceIdiom == .phone
UIDevice.current.userInterfaceIdiom == .unspecified
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)
@unknown default:
// Uh, oh! What could it be?
}
UI_USER_INTERFACE_IDIOM()是一个Objective-C宏，它被定义为:
#define UI_USER_INTERFACE_IDIOM() \ ([[UIDevice currentDevice] respondsToSelector:@selector(userInterfaceIdiom)] ? \ [[UIDevice currentDevice] userInterfaceIdiom] : \ UIUserInterfaceIdiomPhone)
还要注意，即使在使用Objective-C时，UI_USER_INTERFACE_IDIOM()宏也只在针对iOS 3.2及以下时才需要。当部署到iOS 3.2及以上版本时，可以直接使用[UIDevice userInterfaceIdiom]。
如果寇/换:
if UIDevice.current.userInterfaceIdiom == .pad {
// iPad
} else {
// not iPad (iPhone, mac, tv, carPlay, unspecified)
}
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
}
你可以在Swift 5上使用新的方式:
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
}
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