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
从iOS 13开始,UI_USER_INTERFACE_IDIOM已经弃用。如果你的代码仍然是Obj-C,你可以使用以下代码:
if (UIDevice.currentDevice.userInterfaceIdiom == UIUserInterfaceIdiomPad) {
// device is iPad
}
地点:
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
};
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]。
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")
}
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
}
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
你应该使用这个GBDeviceInfo框架或者…
苹果公司这样定义:
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
}
所以对于严格定义的设备可以使用本代码
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")
}