Developing same UI for 3.5, 4.0 (updated 4.7 and 5.5) inches screens in Xcode 5.0.1 (updated xcode 6), no landscape, no iPad and no storyboard

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生来不讨喜
生来不讨喜 2020-12-23 12:37
  1. I have developed app considering 3.5 inch with .xib files and not storyboard.
  2. I am unable to find any tutorial or guide which will help me in designing app si
5条回答
  •  温柔的废话
    2020-12-23 13:14

    SWIFT:

    if(DeviceType.IS_IPHONE_4_OR_LESS)
    {
    //DO THIS 
    }
    //*********************************************************************
        enum UIUserInterfaceIdiom : Int
        {
            case Unspecified
            case Phone
            case Pad
        }
    
        struct ScreenSize
        {
            static let SCREEN_WIDTH         = UIScreen.mainScreen().bounds.size.width
            static let SCREEN_HEIGHT        = UIScreen.mainScreen().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.currentDevice().userInterfaceIdiom == .Phone && ScreenSize.SCREEN_MAX_LENGTH < 568.0
            static let IS_IPHONE_5          = UIDevice.currentDevice().userInterfaceIdiom == .Phone && ScreenSize.SCREEN_MAX_LENGTH == 568.0
            static let IS_IPHONE_6          = UIDevice.currentDevice().userInterfaceIdiom == .Phone && ScreenSize.SCREEN_MAX_LENGTH == 667.0
            static let IS_IPHONE_6P         = UIDevice.currentDevice().userInterfaceIdiom == .Phone && ScreenSize.SCREEN_MAX_LENGTH == 736.0
            static let IS_IPAD              = UIDevice.currentDevice().userInterfaceIdiom == .Pad && ScreenSize.SCREEN_MAX_LENGTH == 1024.0
        }
    

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