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
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
}
其他回答
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
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
}
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
};
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")
}
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")
}