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

我做了一个小函数,把它放在我可以全局使用它的地方,在swift 2.1中工作得很好:

func getColorFromHex(rgbValue:UInt32)->UIColor{
   let red = CGFloat((rgbValue & 0xFF0000) >> 16)/255.0
   let green = CGFloat((rgbValue & 0xFF00) >> 8)/255.0
   let blue = CGFloat(rgbValue & 0xFF)/255.0

   return UIColor(red:red, green:green, blue:blue, alpha:1.0)
}

用法:

getColorFromHex(0xffffff)

其他回答

另一种方法

斯威夫特3.0

为UIColor写一个扩展

// To change the HexaDecimal value to Corresponding Color
extension UIColor
{
    class func uicolorFromHex(_ rgbValue:UInt32, alpha : CGFloat)->UIColor

    {
        let red = CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0
        let green = CGFloat((rgbValue & 0xFF00) >> 8) / 255.0
        let blue = CGFloat(rgbValue & 0xFF) / 255.0
        return UIColor(red:red, green:green, blue:blue, alpha: alpha)
    }
}

你可以像这样用hex直接创建UIColor

let carrot = UIColor.uicolorFromHex(0xe67e22, alpha: 1))

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)

你可以在UIColor上使用这个扩展,它将你的字符串(十六进制,RGBA)转换为UIColor,反之亦然。

extension UIColor {

  //Convert RGBA String to UIColor object
  //"rgbaString" must be separated by space "0.5 0.6 0.7 1.0" 50% of Red 60% of Green 70% of Blue Alpha 100%
  public convenience init?(rgbaString : String){
      self.init(ciColor: CIColor(string: rgbaString))
  }

  //Convert UIColor to RGBA String
  func toRGBAString()-> String {
    var r: CGFloat = 0
    var g: CGFloat = 0
    var b: CGFloat = 0
    var a: CGFloat = 0
    self.getRed(&r, green: &g, blue: &b, alpha: &a)
    return "\(r) \(g) \(b) \(a)"
  }

  //return UIColor from Hexadecimal Color string
  public convenience init?(hexString: String) {  
    let r, g, b, a: CGFloat

    if hexString.hasPrefix("#") {
      let start = hexString.index(hexString.startIndex, offsetBy: 1)
      let hexColor = hexString.substring(from: start)

      if hexColor.characters.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
  }

  // Convert UIColor to Hexadecimal String
  func toHexString() -> String {
    var r: CGFloat = 0
    var g: CGFloat = 0
    var b: CGFloat = 0
    var a: CGFloat = 0
    self.getRed(&r, green: &g, blue: &b, alpha: &a)
    return String(
        format: "%02X%02X%02X",
        Int(r * 0xff),
        Int(g * 0xff),
        Int(b * 0xff))
  }
}

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")

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