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

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")
}

其他回答

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")
}

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]。

如果寇/换:

 if UIDevice.current.userInterfaceIdiom == .pad {
     // iPad
 } else {
     // not iPad (iPhone, mac, tv, carPlay, unspecified)
 }

斯威夫特2. x:

加上别斯拉夫·图拉洛夫的回答,新的iPad Pro可以很容易地找到这一行

检测iPad Pro

struct DeviceType
{
    ...
    static let IS_IPAD_PRO = UIDevice.currentDevice().userInterfaceIdiom == .Pad && ScreenSize.SCREEN_MAX_LENGTH == 1366.0
}

Swift 3(电视和汽车添加):

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            = UIDevice.current.userInterfaceIdiom == .phone
    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          = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 667.0
    static let IS_IPHONE_6P         = UIDevice.current.userInterfaceIdiom == .phone && ScreenSize.SCREEN_MAX_LENGTH == 736.0
    static let IS_IPHONE_7          = IS_IPHONE_6
    static let IS_IPHONE_7P         = IS_IPHONE_6P
    static let IS_IPAD              = UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.SCREEN_MAX_LENGTH == 1024.0
    static let IS_IPAD_PRO_9_7      = IS_IPAD
    static let IS_IPAD_PRO_12_9     = UIDevice.current.userInterfaceIdiom == .pad && ScreenSize.SCREEN_MAX_LENGTH == 1366.0
    static let IS_TV                = UIDevice.current.userInterfaceIdiom == .tv
    static let IS_CAR_PLAY          = UIDevice.current.userInterfaceIdiom == .carPlay
}

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)
    static let iOS10 = (Version.SYS_VERSION_FLOAT >= 10.0 && Version.SYS_VERSION_FLOAT < 11.0)
}

用法:

if DeviceType.IS_IPHONE_7P { print("iPhone 7 plus") }
if DeviceType.IS_IPAD_PRO_9_7 && Version.iOS10 { print("iPad pro 9.7 with iOS 10 version") }

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
}