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
如果寇/换:
if UIDevice.current.userInterfaceIdiom == .pad {
// iPad
} else {
// not iPad (iPhone, mac, tv, carPlay, unspecified)
}
其他回答
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。
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
}
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")
}
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")
}
如果你想检查当前设备是iPad还是iPhone,那么你可以使用这些代码行:
if(UIDevice.currentDevice().userInterfaceIdiom == .Pad){
}else if(UIDevice.currentDevice().userInterfaceIdiom == .Phone){
}