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
对上面的答案做了一些补充，以便返回一个类型而不是字符串值。
我认为这主要是用于UI调整，所以我不认为包括所有的子型号，如iPhone 5s，但这可以很容易地通过添加模型测试扩展到isDevice数组
在Swift 3.1 Xcode 8.3.2中使用物理和模拟器设备进行了测试
实现:
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，其中包括这个和其他一些辅助协议，它是公共的，可以通过Carthage使用。
其他回答
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
}
}
使用
if UIDevice.isPad {
// Do something
}
斯威夫特2. x:
加上别斯拉夫·图拉洛夫的回答，新的iPad Pro可以很容易地找到这一行
检测iPad Pro
struct DeviceType
{
...
static let IS_IPAD_PRO = UIDevice.currentDevice().userInterfaceIdiom == .Pad && ScreenSize.SCREEN_MAX_LENGTH == 1366.0
}
Swift 3(电视和汽车添加):
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") }
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
}
}
}
使用:
if Helper.isIpad {
}
OR
guard Helper.isIpad else {
return
}
由于@user3378170
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")
}
使用UITraitCollection。 iOS trait环境是通过UITraitEnvironment协议的traitCollection属性公开的。以下类采用此协议:
UIScreen ui窗口 ui UIPresentationController UIView
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
}
}
}
下面是你如何使用它:
let modelName = UIDevice.currentDevice().modelName
编辑 对于模拟器，您可以在这里尝试解决方案
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