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

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")
}

其他回答

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
}

如果寇/换:

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

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")
}

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