一个简明的解决方案:

// Assumes input like "#00FF00" (#RRGGBB).
+ (UIColor *)colorFromHexString:(NSString *)hexString {
    unsigned rgbValue = 0;
    NSScanner *scanner = [NSScanner scannerWithString:hexString];
    [scanner setScanLocation:1]; // bypass '#' character
    [scanner scanHexInt:&rgbValue];
    return [UIColor colorWithRed:((rgbValue & 0xFF0000) >> 16)/255.0 green:((rgbValue & 0xFF00) >> 8)/255.0 blue:(rgbValue & 0xFF)/255.0 alpha:1.0];
}

另一个带有alpha的版本

#define UIColorFromRGBA(rgbValue) [UIColor colorWithRed:((float)((rgbValue & 0xFF000000) >> 24))/255.0 green:((float)((rgbValue & 0xFF0000) >> 16))/255.0 blue:((float)((rgbValue & 0xFF00) >> 8 ))/255.0 alpha:((float)((rgbValue & 0xFF))/255.0)]

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。

除了颜色，我还喜欢保证alpha，所以我自己写类别

+ (UIColor *) colorWithHex:(int)color {

    float red = (color & 0xff000000) >> 24;
    float green = (color & 0x00ff0000) >> 16;
    float blue = (color & 0x0000ff00) >> 8;
    float alpha = (color & 0x000000ff);

    return [UIColor colorWithRed:red/255.0 green:green/255.0 blue:blue/255.0 alpha:alpha/255.0];
}

像这样很容易使用

[UIColor colorWithHex:0xFF0000FF]; //Red
[UIColor colorWithHex:0x00FF00FF]; //Green
[UIColor colorWithHex:0x00FF00FF]; //Blue
[UIColor colorWithHex:0x0000007F]; //transparent black

    //UIColorWithHexString

    static UIColor * UIColorWithHexString(NSString *hex) {
        unsigned int rgb = 0;
        [[NSScanner scannerWithString:
          [[hex uppercaseString] stringByTrimmingCharactersInSet:
           [[NSCharacterSet characterSetWithCharactersInString:@"0123456789ABCDEF"] invertedSet]]]
         scanHexInt:&rgb];
        return [UIColor colorWithRed:((CGFloat)((rgb & 0xFF0000) >> 16)) / 255.0
                               green:((CGFloat)((rgb & 0xFF00) >> 8)) / 255.0
                                blue:((CGFloat)(rgb & 0xFF)) / 255.0
                               alpha:1.0];
    }

使用

self.view.backgroundColor = UIColorWithHexString(@"#0F35C0");

在swift中，我用以下方法创建了一个类扩展，将十六进制代码转换为UIColor。

extension UIColor {  
  convenience init(R: CGFloat, G: CGFloat, B: CGFloat, alpha: CGFloat) {
    self.init(red: R/255.0, green: G/255.0, blue: B/255.0, alpha: alpha)
  }

  class func colorWithHex(hex: UInt, alpha: CGFloat) -> UIColor {
    return UIColor(R: CGFloat((hex & 0xFF0000) >> 16), G: CGFloat((hex & 0x00FF00) >> 8), B: CGFloat(hex & 0x0000FF), alpha: alpha)
  }
}