另一个实现允许字符串“FFF”或“FFFFFF”，并使用alpha:

+ (UIColor *) colorFromHexString:(NSString *)hexString alpha: (CGFloat)alpha{
    NSString *cleanString = [hexString stringByReplacingOccurrencesOfString:@"#" withString:@""];
    if([cleanString length] == 3) {
        cleanString = [NSString stringWithFormat:@"%@%@%@%@%@%@",
                       [cleanString substringWithRange:NSMakeRange(0, 1)],[cleanString substringWithRange:NSMakeRange(0, 1)],
                       [cleanString substringWithRange:NSMakeRange(1, 1)],[cleanString substringWithRange:NSMakeRange(1, 1)],
                       [cleanString substringWithRange:NSMakeRange(2, 1)],[cleanString substringWithRange:NSMakeRange(2, 1)]];
    }
    if([cleanString length] == 6) {
        cleanString = [cleanString stringByAppendingString:@"ff"];
    }

    unsigned int baseValue;
    [[NSScanner scannerWithString:cleanString] scanHexInt:&baseValue];

    float red = ((baseValue >> 24) & 0xFF)/255.0f;
    float green = ((baseValue >> 16) & 0xFF)/255.0f;
    float blue = ((baseValue >> 8) & 0xFF)/255.0f;

    return [UIColor colorWithRed:red green:green blue:blue alpha:alpha];
}

大多数发布的解决方案使用了Scanner，但至少在现代Swift中你并不真正需要它。相反，你可以简单地使用UInt init和基数16，然后使用基本的二进制操作来获得UIColor组件:

func stringToColor(color: String) -> UIColor {
    guard let i = UInt(color, radix: 16) else {
        return UIColor.white
    }
    return UIColor(
        red: CGFloat((i & 0xFF0000) >> 16) / 255.0,
        green: CGFloat((i & 0xFF00) >> 8) / 255.0,
        blue: CGFloat(i & 0xFF) / 255.0,
        alpha: 1.0
    )
}

这个解决方案期望输入像“FF00FF”，你可能需要删除前面的哈希符号(#)，如果你的字符串中有一个。

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
    )
  }
}

十六进制初始化

extension UIColor{
public  convenience init(hex : String) {
    var cString:String = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()

    if (cString.hasPrefix("#")) {
        cString.remove(at: cString.startIndex)
    }

    if ((cString.count) != 6) {
        self.init(red: 1, green: 1, blue: 1, alpha: 1)
        return
    }

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

    self.init(
    red: CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0,
    green: CGFloat((rgbValue & 0x00FF00) >> 8) / 255.0,
    blue: CGFloat(rgbValue & 0x0000FF) / 255.0,
    alpha: CGFloat(1.0)
    )
}


//Iniitailization
let myColor = UIColor(hex: "#452b4e")

编码愉快!享受! !

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)

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