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。
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
我是这样做的:
UIDevice.current.model
它显示了设备的名称。
检查是iPad还是iPhone:
if ( UIDevice.current.model.range(of: "iPad") != nil){
print("I AM IPAD")
} else {
print("I AM IPHONE")
}
其他回答
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]。
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
}
Use:
if(Device.IS_IPHONE){
// device is iPhone
}if(Device.IS_IPAD){
// device is iPad (or a Mac running under macOS Catalyst)
}else{
// other
}
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。
如果寇/换:
if UIDevice.current.userInterfaceIdiom == .pad {
// iPad
} else {
// not iPad (iPhone, mac, tv, carPlay, unspecified)
}
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