我如何从十六进制字符串格式创建一个UIColor,如#00FF00?


当前回答

这是另一种选择。

- (UIColor *)colorWithRGBHex:(UInt32)hex
{
    int r = (hex >> 16) & 0xFF;
    int g = (hex >> 8) & 0xFF;
    int b = (hex) & 0xFF;

    return [UIColor colorWithRed:r / 255.0f
                           green:g / 255.0f
                            blue:b / 255.0f
                           alpha:1.0f];
}

其他回答

上面的几个解决方案涉及到一些不必要的nsstring使用。这个UIColor类扩展更简单更快:

+ colorWithHex:(UInt32)hex alpha:(CGFloat)alpha
{
    return [UIColor colorWithRed:((hex & 0xFF0000) >> 16)/255.0
                           green:((hex & 0x00FF00) >> 8)/255.0
                            blue:( hex & 0x0000FF)/255.0
                           alpha:alpha];
}

简单来说:

return [UIColor colorWithHex:0x006400 alpha:1.0]; // HTML darkgreen

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

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")

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]

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。