我的iOS应用程序为UINavigationBar使用了自定义高度,这在新的iPhone X上导致了一些问题。

是否有人已经知道如何通过编程(在Objective-C中)可靠地检测应用程序是否在iPhone X上运行?

编辑:

当然,检查屏幕的大小是可能的,但是,我想知道是否有一些“内置”的方法,如TARGET_OS_IPHONE来检测iOS…

if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone) {
    CGSize screenSize = [[UIScreen mainScreen] bounds].size;
    if (screenSize.height == 812)
        NSLog(@"iPhone X");
}

编辑2:

我不认为,我的问题是一个重复的关联问题。当然,有一些方法可以“测量”当前设备的不同属性,并使用结果来决定使用哪种设备。然而,这并不是我在第一篇编辑中试图强调的问题的实际意义。

真正的问题是:“是否有可能直接检测当前设备是否是iPhone X(例如通过某些SDK功能),还是我必须使用间接测量?”

根据目前给出的答案,我假设答案是“不,没有直接的方法。测量是要走的路。”


当前回答

Swift 3 + 4:

不需要任何设备大小的像素值

//UIApplication+SafeArea.swift

extension UIApplication { 

    static var isDeviceWithSafeArea:Bool {

        if #available(iOS 11.0, *) {
            if let topPadding = shared.keyWindow?.safeAreaInsets.bottom,
                topPadding > 0 {
                return true
            }
        }

        return false
    }
}

例子:

if UIApplication.isDeviceWithSafeArea {
     //e.g. change the frame size height of your UITabBar
}

其他回答

检测设备是否为iPhone X的最佳和最简单的方法是,

https://github.com/stephanheilner/UIDevice-DisplayName

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 , value != 0 else { return identifier}
            return identifier + String(UnicodeScalar(UInt8(value)))}

标识符是iPhone10,3或者iPhone10,6对于iPhone X。

struct ScreenSize {
    static let width = UIScreen.main.bounds.size.width
    static let height = UIScreen.main.bounds.size.height
    static let maxLength = max(ScreenSize.width, ScreenSize.height)
    static let minLength = min(ScreenSize.width, ScreenSize.height)
    static let frame = CGRect(x: 0, y: 0, width: ScreenSize.width, height: ScreenSize.height)
}

struct DeviceType {
    static let iPhone4orLess = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.maxLength < 568.0
    static let iPhone5orSE = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.maxLength == 568.0
    static let iPhone678 = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.maxLength == 667.0
    static let iPhone678p = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.maxLength == 736.0
    static let iPhoneX = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.maxLength == 812.0

    static let IS_IPAD              = UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.maxLength == 1024.0
    static let IS_IPAD_PRO          = UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.maxLength == 1366.0
}

Swift 3 + 4:

不需要任何设备大小的像素值

//UIApplication+SafeArea.swift

extension UIApplication { 

    static var isDeviceWithSafeArea:Bool {

        if #available(iOS 11.0, *) {
            if let topPadding = shared.keyWindow?.safeAreaInsets.bottom,
                topPadding > 0 {
                return true
            }
        }

        return false
    }
}

例子:

if UIApplication.isDeviceWithSafeArea {
     //e.g. change the frame size height of your UITabBar
}
#define IS_IPHONE (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone)
#define IS_IPHONE_X (IS_IPHONE && [[UIScreen mainScreen] bounds].size.height == 812.0f)

所有使用高度的答案都只是故事的一半,原因只有一个。如果你想这样检查当设备朝向是landscapeLeft或landscape ight时检查会失败,因为高度和宽度互换了。

这就是为什么我的解决方案看起来像这样在Swift 4.0:

extension UIScreen {
    ///
    static var isPhoneX: Bool {
        let screenSize = UIScreen.main.bounds.size
        let width = screenSize.width
        let height = screenSize.height
        return min(width, height) == 375 && max(width, height) == 812
    }
}