我如何从十六进制字符串格式创建一个UIColor,如#00FF00?


当前回答

抛光扩展从原来的答案@Tom 请随意在这里更新代码

extension UIColor{
    convenience init (hexString:String) {
        var cleanString:String = hexString.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString

        if (cleanString.hasPrefix("#")) {
            cleanString = cleanString.substringFromIndex(cleanString.startIndex.advancedBy(1))
        }

        if (cleanString.characters.count != 6) {
            self.init()
        }
        else{
            var rgbValue = UInt32()
            let scanner = NSScanner(string: cleanString)
            scanner.scanHexInt(&rgbValue)

            self.init(
                red: CGFloat((rgbValue & 0xFF0000) >> 16)/255.0,
                green: CGFloat((rgbValue & 0xFF00) >> 8)/255.0,
                blue: CGFloat(rgbValue & 0xFF)/255.0,
                alpha: 1.0)
        }
    }
}

其他回答

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。

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")

抛光扩展从原来的答案@Tom 请随意在这里更新代码

extension UIColor{
    convenience init (hexString:String) {
        var cleanString:String = hexString.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString

        if (cleanString.hasPrefix("#")) {
            cleanString = cleanString.substringFromIndex(cleanString.startIndex.advancedBy(1))
        }

        if (cleanString.characters.count != 6) {
            self.init()
        }
        else{
            var rgbValue = UInt32()
            let scanner = NSScanner(string: cleanString)
            scanner.scanHexInt(&rgbValue)

            self.init(
                red: CGFloat((rgbValue & 0xFF0000) >> 16)/255.0,
                green: CGFloat((rgbValue & 0xFF00) >> 8)/255.0,
                blue: CGFloat(rgbValue & 0xFF)/255.0,
                alpha: 1.0)
        }
    }
}

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]

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。