我该如何从十六进制字符串创建UIColor？

Question

我该如何从十六进制字符串创建UIColor？

ioscocoa-touchuicolor

584

我该如何从十六进制字符串格式（例如#00FF00）创建一个UIColor呢？

- Rupesh

Erica还为iOS和OSX开发了一个出色的颜色扩展类别。 - Echilon

这是另一个库：https://github.com/burhanuddin353/TFTColor - Burhanuddin Sunelwala

50个回答

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- Abhishek Verma · Answer 1

你可以创建UIColor的扩展类，如下所示：

extension UIColor {

// MARK: - getColorFromHex /** 此函数将把颜色十六进制代码转换为RGB。*/ }

- parameter color  hex string.

- returns: RGB color code.
*/
class func getColorFromHex(hexString:String)->UIColor{

    var rgbValue : UInt32 = 0
    let scanner:NSScanner =  NSScanner(string: hexString)

    scanner.scanLocation = 1
    scanner.scanHexInt(&rgbValue)

    return UIColor(red: CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0, green: CGFloat((rgbValue & 0x00FF00) >> 8) / 255.0, blue: CGFloat(rgbValue & 0x0000FF) / 255.0, alpha: CGFloat(1.0))
}

}

- Usman · Answer 2

    //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];
    }

使用方法

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

- AntiMoron · Answer 3

对于Swift 2.0+，这段代码对我来说运行得很好。

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))
    }
}

这段代码包括字符串格式检查。例如：

let aColor = UIColor(hexString: "#dadada")!
let failed = UIColor(hexString: "123zzzz")

据我所知，我的代码保持了可失败条件的语义并返回了一个可选值，这对于代码的维护没有任何不利影响。这应该是最好的答案。

- Mike Keskinov · Answer 4

大多数发布的解决方案使用 Scanner，但在现代 Swift 中您实际上不需要它。相反，您可以简单地使用基数为 16 的 UInt init，然后使用基本的二进制操作来获取 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
    )
}

此解决方案期望输入类似于“FF00FF”的内容，如果您的字符串中有前导井号（#），则可能需要将其删除。请注意保留HTML标签。

- Manish Saini · Answer 5

 You Can Get UIColor From String Code Like
   circularSpinner.fillColor = [self getUIColorObjectFromHexString:@"27b8c8" alpha:9];

 //Function For Hex Color Use
    - (unsigned int)intFromHexString:(NSString *)hexStr
    {
        unsigned int hexInt = 0;

        // Create scanner
        NSScanner *scanner = [NSScanner scannerWithString:hexStr];

        // Tell scanner to skip the # character
        [scanner setCharactersToBeSkipped:[NSCharacterSet characterSetWithCharactersInString:@"#"]];

        // Scan hex value
        [scanner scanHexInt:&hexInt];

        return hexInt;
    }




    - (UIColor *)getUIColorObjectFromHexString:(NSString *)hexStr alpha:(CGFloat)alpha
    {
        // Convert hex string to an integer
        unsigned int hexint = [self intFromHexString:hexStr];

        // Create color object, specifying alpha as well
        UIColor *color =
        [UIColor colorWithRed:((CGFloat) ((hexint & 0xFF0000) >> 16))/255
                        green:((CGFloat) ((hexint & 0xFF00) >> 8))/255
                         blue:((CGFloat) (hexint & 0xFF))/255
                        alpha:alpha];

        return color;
    }

    /Function For Hex Color Use
    - (unsigned int)intFromHexString:(NSString *)hexStr
    {
        unsigned int hexInt = 0;

        // Create scanner
        NSScanner *scanner = [NSScanner scannerWithString:hexStr];

        // Tell scanner to skip the # character
        [scanner setCharactersToBeSkipped:[NSCharacterSet characterSetWithCharactersInString:@"#"]];

        // Scan hex value
        [scanner scanHexInt:&hexInt];

        return hexInt;
    }




    - (UIColor *)getUIColorObjectFromHexString:(NSString *)hexStr alpha:(CGFloat)alpha
    {
        // Convert hex string to an integer
        unsigned int hexint = [self intFromHexString:hexStr];

        // Create color object, specifying alpha as well
        UIColor *color =
        [UIColor colorWithRed:((CGFloat) ((hexint & 0xFF0000) >> 16))/255
                        green:((CGFloat) ((hexint & 0xFF00) >> 8))/255
                         blue:((CGFloat) (hexint & 0xFF))/255
                        alpha:alpha];

        return color;
    }

- Lucas Martins Juviniano · Answer 6

我为此创建了一个便利初始化方法：

extension UIColor {
convenience init(hex: String, alpha: CGFloat)
{
    let redH = CGFloat(strtoul(hex.substringToIndex(advance(hex.startIndex,2)), nil, 16))
    let greenH = CGFloat(strtoul(hex.substringWithRange(Range<String.Index>(start: advance(hex.startIndex, 2), end: advance(hex.startIndex, 4))), nil, 16))
    let blueH = CGFloat(strtoul(hex.substringFromIndex(advance(hex.startIndex,4)), nil, 16))

    self.init(red: redH/255, green: greenH/255, blue: blueH/255, alpha: alpha)
}
}

那么您可以在项目的任何地方像这样创建一个UIColor：

UIColor(hex: "ffe3c8", alpha: 1)

希望这可以帮到你...

- AVog · Answer 7

使用任意转换网站将十六进制颜色转换为RGB值（如果您在谷歌中搜索“hex to rgb”，会看到大量的网站）。例如，可以使用这个网站：http://www.rgbtohex.net/hextorgb/。

然后将颜色属性更改为UIColor。例如：

self.profilePicture.layer.borderColor = [UIColor colorWithRed:0 green:167 blue:142 alpha:1.0].CGColor;

十六进制颜色值为：00a78e 转换为RGB：R: 0 G: 167 B: 142

如果您提供的RGB值不在0到1.0之间，您需要将它们除以255。例如：

self.profilePicture.layer.borderColor = [UIColor colorWithRed:83.00/255.0 green:123.00/255.0 blue:53.00/255.0 alpha:1.0].CGColor;

- Rajesh Loganathan · Answer 8

尝试这个：这段代码将从你的十六进制颜色字符串返回 UIColor。

- (UIColor*)colorWithHexString:(NSString*)hex  
{  
    NSString *cString = [[hex stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceAndNewlineCharacterSet]] uppercaseString];  

    // String should be 6 or 8 characters  
    if ([cString length] < 6) return [UIColor grayColor];  

    // strip 0X if it appears  
    if ([cString hasPrefix:@"0X"]) cString = [cString substringFromIndex:2];  

    if ([cString length] != 6) return  [UIColor grayColor];  

    // Separate into r, g, b substrings  
    NSRange range;  
    range.location = 0;  
    range.length = 2;  
    NSString *rString = [cString substringWithRange:range];  

    range.location = 2;  
    NSString *gString = [cString substringWithRange:range];  

    range.location = 4;  
    NSString *bString = [cString substringWithRange:range];  

    // Scan values  
    unsigned int r, g, b;  
    [[NSScanner scannerWithString:rString] scanHexInt:&r];  
    [[NSScanner scannerWithString:gString] scanHexInt:&g];  
    [[NSScanner scannerWithString:bString] scanHexInt:&b];  

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

- Martin Zikmund · Answer 9

在Xamarin.iOS中，您可以使用以下内容代替宏：

public UIColor UIColorFromHexValue(int value, float alpha = 1f) =>
    UIColor.FromRGBA(
        ((value & 0xFF0000) >> 16) / 255.0f,
        ((value & 0x00FF00) >> 16) / 255.0f,
        ((value & 0x0000FF) >> 16) / 255.0f,
        alpha);

使用方式类似：

label.TextColor = UIColorFromHexValue(0xBC1128);

- Alvin George · Answer 10

Swift 2.0:

将此方法添加到VC或UIColor的扩展中。

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))
    }

使用方法:

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

     OR

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