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

我从这个回答中总结了一些想法,并针对iOS 13和Swift 5进行了更新。

extension UIColor {
  
  convenience init(_ hex: String, alpha: CGFloat = 1.0) {
    var cString = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
    
    if cString.hasPrefix("#") { cString.removeFirst() }
    
    if cString.count != 6 {
      self.init("ff0000") // return red color for wrong hex input
      return
    }
    
    var rgbValue: UInt64 = 0
    Scanner(string: cString).scanHexInt64(&rgbValue)
    
    self.init(red: CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0,
              green: CGFloat((rgbValue & 0x00FF00) >> 8) / 255.0,
              blue: CGFloat(rgbValue & 0x0000FF) / 255.0,
              alpha: alpha)
  }

}

然后你可以这样使用它:

UIColor("#ff0000") // with #
UIColor("ff0000")  // without #
UIColor("ff0000", alpha: 0.5) // using optional alpha value

其他回答

斯威夫特2.3: 用户界面颜色扩展。我认为这样更简单。

extension UIColor {
    static func colorFromHex(hexString: String, alpha: CGFloat = 1) -> UIColor {
        //checking if hex has 7 characters or not including '#'
        if hexString.characters.count < 7 {
            return UIColor.whiteColor()
        }
        //string by removing hash
        let hexStringWithoutHash = hexString.substringFromIndex(hexString.startIndex.advancedBy(1))

        //I am extracting three parts of hex color Red (first 2 characters), Green (middle 2 characters), Blue (last two characters)
        let eachColor = [
            hexStringWithoutHash.substringWithRange(hexStringWithoutHash.startIndex...hexStringWithoutHash.startIndex.advancedBy(1)),
            hexStringWithoutHash.substringWithRange(hexStringWithoutHash.startIndex.advancedBy(2)...hexStringWithoutHash.startIndex.advancedBy(3)),
            hexStringWithoutHash.substringWithRange(hexStringWithoutHash.startIndex.advancedBy(4)...hexStringWithoutHash.startIndex.advancedBy(5))]

        let hexForEach = eachColor.map {CGFloat(Int($0, radix: 16) ?? 0)} //radix is base of numeric system you want to convert to, Hexadecimal has base 16

        //return the color by making color
        return UIColor(red: hexForEach[0] / 255, green: hexForEach[1] / 255, blue: hexForEach[2] / 255, alpha: alpha)
    }
}

用法:

let color = UIColor.colorFromHex("#25ac09")

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)
    )
}

我从这个回答中总结了一些想法,并针对iOS 13和Swift 5进行了更新。

extension UIColor {
  
  convenience init(_ hex: String, alpha: CGFloat = 1.0) {
    var cString = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
    
    if cString.hasPrefix("#") { cString.removeFirst() }
    
    if cString.count != 6 {
      self.init("ff0000") // return red color for wrong hex input
      return
    }
    
    var rgbValue: UInt64 = 0
    Scanner(string: cString).scanHexInt64(&rgbValue)
    
    self.init(red: CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0,
              green: CGFloat((rgbValue & 0x00FF00) >> 8) / 255.0,
              blue: CGFloat(rgbValue & 0x0000FF) / 255.0,
              alpha: alpha)
  }

}

然后你可以这样使用它:

UIColor("#ff0000") // with #
UIColor("ff0000")  // without #
UIColor("ff0000", alpha: 0.5) // using optional alpha value

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:

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)