我如何从十六进制字符串格式创建一个UIColor,如#00FF00?


当前回答

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。

其他回答

斯威夫特版本:

extension UIColor {
    convenience init?(var hex: String) {
        hex = hex.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString
        hex = (hex.hasPrefix("#")) ? hex.substringFromIndex(advance(hex.startIndex, 1)) : hex

        var value: UInt32 = 0
        if NSScanner(string: hex).scanHexInt(&value) {
            if count(hex) == 8 {
                self.init(red: CGFloat((value & 0xFF000000) >> 24) / 255.0,
                    green: CGFloat((value & 0x00FF0000) >> 16) / 255.0,
                    blue: CGFloat((value & 0x0000FF00) >> 8) / 255.0,
                    alpha: CGFloat((value & 0x000000FF)) / 255.0)
                return
            } else if count(hex) == 6 {
                self.init(red: CGFloat((value & 0xFF0000) >> 16) / 255.0,
                    green: CGFloat((value & 0x00FF00) >> 8) / 255.0,
                    blue: CGFloat(value & 0x0000FF) / 255.0,
                    alpha: 1.0)
                return
            }
        }
        self.init()
        return nil
    }
}

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。

Swift 2.0 - Xcode 7.2

为UIColor添加扩展。

文件-新建- Swift文件-命名。添加以下内容。

extension UIColor {
    convenience init(hexString:String) {
        let hexString:NSString = hexString.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
        let scanner            = NSScanner(string: hexString as String)
        if (hexString.hasPrefix("#")) {
            scanner.scanLocation = 1
        }

        var color:UInt32 = 0
        scanner.scanHexInt(&color)

        let mask = 0x000000FF
        let r = Int(color >> 16) & mask
        let g = Int(color >> 8) & mask
        let b = Int(color) & mask

        let red   = CGFloat(r) / 255.0
        let green = CGFloat(g) / 255.0
        let blue  = CGFloat(b) / 255.0
        self.init(red:red, green:green, blue:blue, alpha:1)
    }

    func toHexString() -> String {
        var r:CGFloat = 0
        var g:CGFloat = 0
        var b:CGFloat = 0
        var a:CGFloat = 0
        getRed(&r, green: &g, blue: &b, alpha: &a)
        let rgb:Int = (Int)(r*255)<<16 | (Int)(g*255)<<8 | (Int)(b*255)<<0
        return NSString(format:"#%06x", rgb) as String
    }        
}

用法:

Ex. Setting Button's color from hexCode.
    override func viewWillAppear(animated: Bool) {
        loginButton.tintColor = UIColor(hexString: " hex code here ")
}

Ex. Converting Button's current color to hex Code.

    override func viewWillAppear(animated: Bool) {
        let hexString = loginButton.tintColor.toHexString()
        print("HEX STRING: \(hexString)")

    }

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)

斯威夫特4

你可以像这样在扩展中创建一个非常方便的构造函数:

extension UIColor {
    convenience init(hexString: String, alpha: CGFloat = 1.0) {
        var hexInt: UInt32 = 0
        let scanner = Scanner(string: hexString)
        scanner.charactersToBeSkipped = CharacterSet(charactersIn: "#")
        scanner.scanHexInt32(&hexInt)

        let red = CGFloat((hexInt & 0xff0000) >> 16) / 255.0
        let green = CGFloat((hexInt & 0xff00) >> 8) / 255.0
        let blue = CGFloat((hexInt & 0xff) >> 0) / 255.0
        let alpha = alpha

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

以后再用

let color = UIColor(hexString: "#AABBCCDD")