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

支持7十六进制颜色类型

有7种十六进制颜色格式:""#FF0000","0xFF0000", "FF0000", "F00", "red", 0x00FF00, 16711935

NSColorParser.nsColor("#FF0000",1)//red nsColor
NSColorParser.nsColor("FF0",1)//red nsColor
NSColorParser.nsColor("0xFF0000",1)//red nsColor
NSColorParser.nsColor("#FF0000",1)//red nsColor
NSColorParser.nsColor("FF0000",1)//red nsColor
NSColorParser.nsColor(0xFF0000,1)//red nsColor
NSColorParser.nsColor(16711935,1)//red nsColor

注意:这不是一个“单文件解决方案”,有一些依赖关系,但查找它们可能比从头开始研究更快。

https://github.com/eonist/swift-utils/blob/2882002682c4d2a3dc7cb3045c45f66ed59d566d/geom/color/NSColorParser.swift

永久链接: https://github.com/eonist/Element/wiki/Progress#supporting-7-hex-color-types

其他回答

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)
extension UIColor {
    public convenience init?(hex: String) {
        let r, g, b, a: CGFloat

        if hex.hasPrefix("#") {
            let start = hex.index(hex.startIndex, offsetBy: 1)
            let hexColor = String(hex[start...])

            if hexColor.count == 8 {
                let scanner = Scanner(string: hexColor)
                var hexNumber: UInt64 = 0

                if scanner.scanHexInt64(&hexNumber) {
                    r = CGFloat((hexNumber & 0xff000000) >> 24) / 255
                    g = CGFloat((hexNumber & 0x00ff0000) >> 16) / 255
                    b = CGFloat((hexNumber & 0x0000ff00) >> 8) / 255
                    a = CGFloat(hexNumber & 0x000000ff) / 255

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

        return nil
    }
}

用法:

let white = UIColor(hex: "#ffffff")

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)

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)

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