下面是Swift 1.2版本，作为UIColor的扩展。这允许你这样做

let redColor = UIColor(hex: "#FF0000")

我觉得这是最自然的做法。

extension UIColor {
  // Initialiser for strings of format '#_RED_GREEN_BLUE_'
  convenience init(hex: String) {
    let redRange    = Range<String.Index>(start: hex.startIndex.advancedBy(1), end: hex.startIndex.advancedBy(3))
    let greenRange  = Range<String.Index>(start: hex.startIndex.advancedBy(3), end: hex.startIndex.advancedBy(5))
    let blueRange   = Range<String.Index>(start: hex.startIndex.advancedBy(5), end: hex.startIndex.advancedBy(7))

    var red     : UInt32 = 0
    var green   : UInt32 = 0
    var blue    : UInt32 = 0

    NSScanner(string: hex.substringWithRange(redRange)).scanHexInt(&red)
    NSScanner(string: hex.substringWithRange(greenRange)).scanHexInt(&green)
    NSScanner(string: hex.substringWithRange(blueRange)).scanHexInt(&blue)

    self.init(
      red: CGFloat(red) / 255,
      green: CGFloat(green) / 255,
      blue: CGFloat(blue) / 255,
      alpha: 1
    )
  }
}

为UIColor创建优雅的扩展:

extension UIColor {

convenience init(string: String) {

        var uppercasedString = string.uppercased()
        uppercasedString.remove(at: string.startIndex)

        var rgbValue: UInt32 = 0
        Scanner(string: uppercasedString).scanHexInt32(&rgbValue)

        let red = CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0
        let green = CGFloat((rgbValue & 0x00FF00) >> 8) / 255.0
        let blue = CGFloat(rgbValue & 0x0000FF) / 255.0

        self.init(red: red, green: green, blue: blue, alpha: 1)
    }
}

创建红色:

let red = UIColor(string: "#ff0000") 

有一个很好的UIColor类别，其中有许多功能。

用法:

textView.textColor = [UIColor colorWithHexString:textColorHex];
NSLog(@"Text Color Hex: %@", textColorHex);

其中textColorHex有一个形式的@“FFFFFF”没有#符号。

据我所知，没有从十六进制字符串到UIColor(或CGColor)的内置转换。但是，您可以很容易地为此目的编写几个函数—例如，参见iphone开发访问uicolor组件

swift 2.0+。 这段代码对我来说很好。

extension UIColor {
    /// UIColor(hexString: "#cc0000")
    internal convenience init?(hexString:String) {
        guard hexString.characters[hexString.startIndex] == Character("#") else {
            return nil
        }
        guard hexString.characters.count == "#000000".characters.count else {
            return nil
        }
        let digits = hexString.substringFromIndex(hexString.startIndex.advancedBy(1))
        guard Int(digits,radix:16) != nil else{
            return nil
        }
        let red = digits.substringToIndex(digits.startIndex.advancedBy(2))
        let green = digits.substringWithRange(Range<String.Index>(start: digits.startIndex.advancedBy(2),
            end: digits.startIndex.advancedBy(4)))
        let blue = digits.substringWithRange(Range<String.Index>(start:digits.startIndex.advancedBy(4),
            end:digits.startIndex.advancedBy(6)))
        let redf = CGFloat(Double(Int(red, radix:16)!) / 255.0)
        let greenf = CGFloat(Double(Int(green, radix:16)!) / 255.0)
        let bluef = CGFloat(Double(Int(blue, radix:16)!) / 255.0)
        self.init(red: redf, green: greenf, blue: bluef, alpha: CGFloat(1.0))
    }
}

此代码包括字符串格式检查。 如。

let aColor = UIColor(hexString: "#dadada")!
let failed = UIColor(hexString: "123zzzz")

据我所知，我的代码在维护可失败条件的语义和返回可选值方面没有任何缺点。这应该是最好的答案。

Swift等价于@Tom的答案，尽管接收RGBA Int值以支持透明度:

func colorWithHex(aHex: UInt) -> UIColor
{
    return UIColor(red: CGFloat((aHex & 0xFF000000) >> 24) / 255,
        green: CGFloat((aHex & 0x00FF0000) >> 16) / 255,
        blue: CGFloat((aHex & 0x0000FF00) >> 8) / 255,
        alpha: CGFloat((aHex & 0x000000FF) >> 0) / 255)
}

//usage
var color = colorWithHex(0x7F00FFFF)

如果你想从string中使用它，你可以使用strtoul:

var hexString = "0x7F00FFFF"

let num = strtoul(hexString, nil, 16)

var colorFromString = colorWithHex(num)