我如何从十六进制字符串格式创建一个UIColor,如#00FF00?


当前回答

我为它创建了一个便利的init:

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)
}
}

然后你可以在项目的任何地方创建一个UIColor,就像这样:

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

希望这对你有所帮助……

其他回答

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]

另一个实现允许字符串“FFF”或“FFFFFF”,并使用alpha:

+ (UIColor *) colorFromHexString:(NSString *)hexString alpha: (CGFloat)alpha{
    NSString *cleanString = [hexString stringByReplacingOccurrencesOfString:@"#" withString:@""];
    if([cleanString length] == 3) {
        cleanString = [NSString stringWithFormat:@"%@%@%@%@%@%@",
                       [cleanString substringWithRange:NSMakeRange(0, 1)],[cleanString substringWithRange:NSMakeRange(0, 1)],
                       [cleanString substringWithRange:NSMakeRange(1, 1)],[cleanString substringWithRange:NSMakeRange(1, 1)],
                       [cleanString substringWithRange:NSMakeRange(2, 1)],[cleanString substringWithRange:NSMakeRange(2, 1)]];
    }
    if([cleanString length] == 6) {
        cleanString = [cleanString stringByAppendingString:@"ff"];
    }

    unsigned int baseValue;
    [[NSScanner scannerWithString:cleanString] scanHexInt:&baseValue];

    float red = ((baseValue >> 24) & 0xFF)/255.0f;
    float green = ((baseValue >> 16) & 0xFF)/255.0f;
    float blue = ((baseValue >> 8) & 0xFF)/255.0f;

    return [UIColor colorWithRed:red green:green blue:blue alpha:alpha];
}

使用Xcode的原生颜色文字功能来轻松地添加十六进制颜色。

在你的代码中输入颜色文字,然后让Xcode自动完成剩下的工作。

颜色选择界面将允许你粘贴十六进制颜色:#FF9300

宏的git差异将显示RGB值而不是十六进制值:

let orange = #colorLiteral(red: 1, green: 0.5763723254, blue: 0, alpha: 1)

但它仍然是一种简单的方法来粘贴十六进制没有任何第三方工具或扩展。

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")

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。