我试图在Swift中使用十六进制颜色值,而不是UIColor允许您使用的少数标准值,但我不知道如何做到这一点。

示例:我如何使用#ffffff作为颜色?


当前回答

Swift 5:你可以在Xcode中创建颜色,如下图所示:

您应该命名颜色,因为您通过名称引用了颜色。如图2所示:

其他回答

用户界面颜色:

extension UIColor {

    convenience init(hex: Int) {
        let components = (
            R: CGFloat((hex >> 16) & 0xff) / 255,
            G: CGFloat((hex >> 08) & 0xff) / 255,
            B: CGFloat((hex >> 00) & 0xff) / 255
        )
        self.init(red: components.R, green: components.G, blue: components.B, alpha: 1)
    }

}

CGColor:

extension CGColor {

    class func colorWithHex(hex: Int) -> CGColorRef {

        return UIColor(hex: hex).CGColor

    }

}

使用

let purple = UIColor(hex: 0xAB47BC)
extension UIColor {

    convenience init(r: CGFloat, g: CGFloat, b: CGFloat, a: CGFloat = 1) {
        self.init(red: r/255, green: g/255, blue: b/255, alpha: a)
    }

    convenience init(hex: Int, alpha: CGFloat = 1) {
        self.init(r: CGFloat((hex >> 16) & 0xff), g: CGFloat((hex >> 08) & 0xff), b: CGFloat((hex >> 00) & 0xff), a: alpha)
    }
}

RGBA版本Swift 3/4

我喜欢卢卡的回答,因为我认为它是最优雅的。

然而,我不希望我的颜色指定在ARGB。我宁愿RGBA +,我也需要在处理字符串的情况下,为每个频道指定1个字符“#FFFA”。

这个版本还增加了错误抛出+剥离'#'字符如果它包含在字符串中。 这是我修改后的Swift表格。

public enum ColourParsingError: Error
{
    
    case invalidInput(String)
}
extension UIColor {
    public convenience init(hexString: String) throws
    {
        let hexString = hexString.replacingOccurrences(of: "#", with: "")
        let hex = hexString.trimmingCharacters(in:NSCharacterSet.alphanumerics.inverted)
        var int = UInt32()
        Scanner(string: hex).scanHexInt32(&int)
        let a, r, g, b: UInt32
        switch hex.count 
        {
        case 3: // RGB (12-bit)
            (r, g, b,a) = ((int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17,255)
        //iCSS specification in the form of #F0FA
        case 4: // RGB (24-bit)
            (r, g, b,a) = ((int >> 12) * 17, (int >> 8 & 0xF) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17)
        case 6: // RGB (24-bit)
            (r, g, b, a) = (int >> 16, int >> 8 & 0xFF, int & 0xFF,255)
        case 8: // ARGB (32-bit)
            (r, g, b, a) = (int >> 24, int >> 16 & 0xFF, int >> 8 & 0xFF, int & 0xFF)
        default:
            throw ColourParsingError.invalidInput("String is not a valid hex colour string: \(hexString)")
        }
        self.init(red: CGFloat(r) / 255, green: CGFloat(g) / 255, blue: CGFloat(b) / 255, alpha: CGFloat(a) / 255)
    }
}

这个答案展示了如何在Obj-C中实现。这座桥是要用的

let rgbValue = 0xFFEEDD
let r = Float((rgbValue & 0xFF0000) >> 16)/255.0
let g = Float((rgbValue & 0xFF00) >> 8)/255.0
let b = Float((rgbValue & 0xFF))/255.0
self.backgroundColor = UIColor(red:r, green: g, blue: b, alpha: 1.0)

只是对第一个答案的一些补充

(还没有检查alpha,可能需要添加一个if next > 0xffffff):

extension UIColor {

struct COLORS_HEX {
    static let Primary = 0xffffff
    static let PrimaryDark = 0x000000
    static let Accent = 0xe89549
    static let AccentDark = 0xe27b2a
    static let TextWhiteSemiTransparent = 0x80ffffff
}

convenience init(red: Int, green: Int, blue: Int, alphaH: Int) {
    assert(red >= 0 && red <= 255, "Invalid red component")
    assert(green >= 0 && green <= 255, "Invalid green component")
    assert(blue >= 0 && blue <= 255, "Invalid blue component")
    assert(alphaH >= 0 && alphaH <= 255, "Invalid alpha component")

    self.init(red: CGFloat(red) / 255.0, green: CGFloat(green) / 255.0, blue: CGFloat(blue) / 255.0, alpha: CGFloat(alphaH) / 255.0)
}

convenience init(netHex:Int) {
    self.init(red:(netHex >> 16) & 0xff, green:(netHex >> 8) & 0xff, blue:netHex & 0xff, alphaH: (netHex >> 24) & 0xff)
}

}