我如何从十六进制字符串格式创建一个UIColor,如#00FF00?


当前回答

你可以使用这个库

https://github.com/burhanuddin353/TFTColor

斯威夫特

UIColor.colorWithRGB(hexString: "FF34AE" alpha: 1.0)

objective - c

[UIColor colorWithRGBHexString:@"FF34AE" alpha:1.0f]

其他回答

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。

有cocoapod支持,这很好

https://github.com/mRs-/HexColors

// with hash
NSColor *colorWithHex = [NSColor colorWithHexString:@"#ff8942" alpha:1];

// wihtout hash
NSColor *secondColorWithHex = [NSColor colorWithHexString:@"ff8942" alpha:1];

// short handling
NSColor *shortColorWithHex = [NSColor colorWithHexString:@"fff" alpha:1]

张贴参考网站我刚刚找到。 它做了所有的脏工作,从HEX或RGB开始,在ObjC, Swift和Xamarin中打印出代码。

https://www.uicolor.xyz/#/hex-to-ui

斯威夫特2.0:

将此方法添加到VC或扩展UIColor。

func colorWithHexString (hex:String) -> UIColor {
        var cString:String = hex.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString

        if (cString.hasPrefix("#")) {
            cString = (cString as NSString).substringFromIndex(1)
        }

        if (cString.characters.count != 6) {
            return UIColor.grayColor()
        }

        let rString = (cString as NSString).substringToIndex(2)
        let gString = ((cString as NSString).substringFromIndex(2) as NSString).substringToIndex(2)
        let bString = ((cString as NSString).substringFromIndex(4) as NSString).substringToIndex(2)

        var r:CUnsignedInt = 0, g:CUnsignedInt = 0, b:CUnsignedInt = 0;
        NSScanner(string: rString).scanHexInt(&r)
        NSScanner(string: gString).scanHexInt(&g)
        NSScanner(string: bString).scanHexInt(&b)


        return UIColor(red: CGFloat(r) / 255.0, green: CGFloat(g) / 255.0, blue: CGFloat(b) / 255.0, alpha: CGFloat(1))
    }

用法:

    loginButton.tintColor = self.colorWithHexString("#be1337")

     OR

    let hexColor = self.colorWithHexString("#be1337")

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)