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

这是我用的。适用于6和8字符的颜色字符串,带或不带#符号。在发布时默认为黑色,在调试时用无效字符串初始化时崩溃。

extension UIColor {
    public convenience init(hex: String) {
        var r: CGFloat = 0
        var g: CGFloat = 0
        var b: CGFloat = 0
        var a: CGFloat = 1

        let hexColor = hex.replacingOccurrences(of: "#", with: "")
        let scanner = Scanner(string: hexColor)
        var hexNumber: UInt64 = 0
        var valid = false

        if scanner.scanHexInt64(&hexNumber) {
            if hexColor.count == 8 {
                r = CGFloat((hexNumber & 0xff000000) >> 24) / 255
                g = CGFloat((hexNumber & 0x00ff0000) >> 16) / 255
                b = CGFloat((hexNumber & 0x0000ff00) >> 8) / 255
                a = CGFloat(hexNumber & 0x000000ff) / 255
                valid = true
            }
            else if hexColor.count == 6 {
                r = CGFloat((hexNumber & 0xff0000) >> 16) / 255
                g = CGFloat((hexNumber & 0x00ff00) >> 8) / 255
                b = CGFloat(hexNumber & 0x0000ff) / 255
                valid = true
            }
        }

        #if DEBUG
            assert(valid, "UIColor initialized with invalid hex string")
        #endif

        self.init(red: r, green: g, blue: b, alpha: a)
    }
}

用法:

UIColor(hex: "#75CC83FF")
UIColor(hex: "75CC83FF")
UIColor(hex: "#75CC83")
UIColor(hex: "75CC83")

其他回答

Xcode 13.2.1, M1, Swift 5.5

我们可以在ColorLiterals中使用Hex

输入#colorLiteral(在Xcode中,这将触发并修复与ColorLiterals相关的错误

然后点击其他

然后选择RGB滑块,你现在可以看到十六进制面板

iOS 14, SwiftUI 2.0, swift 5.1, Xcode beta12

extension Color {
  static func hexColour(hexValue:UInt32)->Color
    {
      let red = Double((hexValue & 0xFF0000) >> 16) / 255.0
      let green = Double((hexValue & 0xFF00) >> 8) / 255.0
      let blue = Double(hexValue & 0xFF) / 255.0
      return Color(red:red, green:green, blue:blue)
    }
}

用十六进制数表示

let red = Color.hexColour(hexValue: 0xFF0000)

斯威夫特2.0:

在viewDidLoad ()

 var viewColor:UIColor
    viewColor = UIColor()
    let colorInt:UInt
    colorInt = 0x000000
    viewColor = UIColorFromRGB(colorInt)
    self.View.backgroundColor=viewColor



func UIColorFromRGB(rgbValue: UInt) -> UIColor {
    return UIColor(
        red: CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0,
        green: CGFloat((rgbValue & 0x00FF00) >> 8) / 255.0,
        blue: CGFloat(rgbValue & 0x0000FF) / 255.0,
        alpha: CGFloat(1.0)
    )
}

在Swift 2.0和Xcode 7.0.1中,你可以创建这个函数:

    // Creates a UIColor from a Hex string.
    func colorWithHexString (hex:String) -> UIColor {
        var cString:String = hex.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString

        if (cString.hasPrefix("#")) {
            cString = (cString as NSString).substringFromIndex(1)
        }

        if (cString.characters.count != 6) {
            return UIColor.grayColor()
        }

        let rString = (cString as NSString).substringToIndex(2)
        let gString = ((cString as NSString).substringFromIndex(2) as NSString).substringToIndex(2)
        let bString = ((cString as NSString).substringFromIndex(4) as NSString).substringToIndex(2)

        var r:CUnsignedInt = 0, g:CUnsignedInt = 0, b:CUnsignedInt = 0;
        NSScanner(string: rString).scanHexInt(&r)
        NSScanner(string: gString).scanHexInt(&g)
        NSScanner(string: bString).scanHexInt(&b)


        return UIColor(red: CGFloat(r) / 255.0, green: CGFloat(g) / 255.0, blue: CGFloat(b) / 255.0, alpha: CGFloat(1))
    }

然后这样使用它:

let color1 = colorWithHexString("#1F437C")

Swift 4更新

func colorWithHexString (hex:String) -> UIColor {

    var cString = hex.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines).uppercased()

    if (cString.hasPrefix("#")) {
        cString = (cString as NSString).substring(from: 1)
    }

    if (cString.characters.count != 6) {
        return UIColor.gray
    }

    let rString = (cString as NSString).substring(to: 2)
    let gString = ((cString as NSString).substring(from: 2) as NSString).substring(to: 2)
    let bString = ((cString as NSString).substring(from: 4) as NSString).substring(to: 2)

    var r:CUnsignedInt = 0, g:CUnsignedInt = 0, b:CUnsignedInt = 0;
    Scanner(string: rString).scanHexInt32(&r)
    Scanner(string: gString).scanHexInt32(&g)
    Scanner(string: bString).scanHexInt32(&b)


    return UIColor(red: CGFloat(r) / 255.0, green: CGFloat(g) / 255.0, blue: CGFloat(b) / 255.0, alpha: CGFloat(1))
}

最新swift3版本

        extension UIColor {
convenience init(hexString: String) {
    let hex = hexString.trimmingCharacters(in: CharacterSet.alphanumerics.inverted)
    var int = UInt32()
    Scanner(string: hex).scanHexInt32(&int)
    let a, r, g, b: UInt32
    switch hex.characters.count {
    case 3: // RGB (12-bit)
        (a, r, g, b) = (255, (int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17)
    case 6: // RGB (24-bit)
        (a, r, g, b) = (255, int >> 16, int >> 8 & 0xFF, int & 0xFF)
    case 8: // ARGB (32-bit)
        (a, r, g, b) = (int >> 24, int >> 16 & 0xFF, int >> 8 & 0xFF, int & 0xFF)
    default:
        (a, r, g, b) = (255, 0, 0, 0)
    }
      self.init(red: CGFloat(r) / 255, green: CGFloat(g) / 255, blue:      CGFloat(b) / 255, alpha: CGFloat(a) / 255)
}
}

在你的类或任何你把hexcolor转换为uicolor的地方使用这种方法

             let color1 = UIColor(hexString: "#FF323232")