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

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")

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")

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。

Swift 5, iOS 14

convenience init(hex: String, alpha: CGFloat = 1.0) {
    var hexFormatted: String = hex.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines).uppercased()
    
    if hexFormatted.hasPrefix("#") {
        hexFormatted = String(hexFormatted.dropFirst())
    }
    
    assert(hexFormatted.count == 6, "Invalid hex code used.")
    
    var rgbValue: UInt64 = 0
    Scanner(string: hexFormatted).scanHexInt64(&rgbValue)
    
    self.init(red: CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0,
              green: CGFloat((rgbValue & 0x00FF00) >> 8) / 255.0,
              blue: CGFloat(rgbValue & 0x0000FF) / 255.0,
              alpha: alpha)
}

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