我试图在Swift中使用十六进制颜色值,而不是UIColor允许您使用的少数标准值,但我不知道如何做到这一点。

示例:我如何使用#ffffff作为颜色?


当前回答

Xcode 13.2.1, M1, Swift 5.5

我们可以在ColorLiterals中使用Hex

输入#colorLiteral(在Xcode中,这将触发并修复与ColorLiterals相关的错误

然后点击其他

然后选择RGB滑块,你现在可以看到十六进制面板

其他回答

这是一个接受十六进制字符串并返回UIColor的函数。 (你可以输入十六进制字符串格式:#ffffff或ffffff)

用法:

var color1 = hexStringToUIColor("#d3d3d3")

Swift 5:(Swift 4+)

func hexStringToUIColor (hex:String) -> UIColor {
    var cString:String = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()

    if (cString.hasPrefix("#")) {
        cString.remove(at: cString.startIndex)
    }

    if ((cString.count) != 6) {
        return UIColor.gray
    }

    var rgbValue:UInt64 = 0
    Scanner(string: cString).scanHexInt64(&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)
    )
}

斯威夫特3:

func hexStringToUIColor (hex:String) -> UIColor {
    var cString:String = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()

    if (cString.hasPrefix("#")) {
        cString.remove(at: cString.startIndex)
    }

    if ((cString.characters.count) != 6) {
        return UIColor.gray
    }

    var rgbValue:UInt32 = 0
    Scanner(string: cString).scanHexInt32(&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)
    )
}

斯威夫特2:

func hexStringToUIColor (hex:String) -> UIColor {
    var cString:String = hex.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet() as NSCharacterSet).uppercaseString

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

    if ((cString.characters.count) != 6) {
      return UIColor.grayColor()
    }

    var rgbValue:UInt32 = 0
    NSScanner(string: cString).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)
    )
}


沙德源代码:/ gist: de147c42d7b3063ef7bc

编辑:更新代码。谢谢,Hlung, jaytrixz, Ahmad F, Kegham K和Adam Waite!

斯威夫特2.0:

在viewDidLoad ()

 var viewColor:UIColor
    viewColor = UIColor()
    let colorInt:UInt
    colorInt = 0x000000
    viewColor = UIColorFromRGB(colorInt)
    self.View.backgroundColor=viewColor



func UIColorFromRGB(rgbValue: UInt) -> UIColor {
    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)
    )
}

Swift 5:你可以在Xcode中创建颜色,如下图所示:

您应该命名颜色,因为您通过名称引用了颜色。如图2所示:

修复了“'scanHexInt32' was deprecated in iOS 13.0”的警告。

示例应该可以在Swift2.2及以上版本(Swift2. exe)上运行。x, Swift3。x, Swift4。x, Swift5.x):

extension UIColor {

    // hex sample: 0xf43737
    convenience init(_ hex: Int, alpha: Double = 1.0) {
        self.init(red: CGFloat((hex >> 16) & 0xFF) / 255.0, green: CGFloat((hex >> 8) & 0xFF) / 255.0, blue: CGFloat((hex) & 0xFF) / 255.0, alpha: CGFloat(255 * alpha) / 255)
    }

    convenience init(_ hexString: String, alpha: Double = 1.0) {
        let hex = hexString.trimmingCharacters(in: CharacterSet.alphanumerics.inverted)
        var int = UInt64()
        Scanner(string: hex).scanHexInt64(&int)

        let r, g, b: UInt64
        switch hex.count {
        case 3: // RGB (12-bit)
            (r, g, b) = ((int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17)
        case 6: // RGB (24-bit)
            (r, g, b) = (int >> 16, int >> 8 & 0xFF, int & 0xFF)
        default:
            (r, g, b) = (1, 1, 0)
        }

        self.init(red: CGFloat(r) / 255, green: CGFloat(g) / 255, blue: CGFloat(b) / 255, alpha: CGFloat(255 * alpha) / 255)
    }

    convenience init(r: CGFloat, g: CGFloat, b: CGFloat, a: CGFloat = 1) {
        self.init(red: (r / 255), green: (g / 255), blue: (b / 255), alpha: a)
    }
}

像下面这样使用它们:

UIColor(0xF54A45)
UIColor(0xF54A45, alpha: 0.7)
UIColor("#f44")
UIColor("#f44", alpha: 0.7)
UIColor("#F54A45")
UIColor("#F54A45", alpha: 0.7)
UIColor("F54A45")
UIColor("F54A45", alpha: 0.7)
UIColor(r: 245.0, g: 73, b: 69)
UIColor(r: 245.0, g: 73, b: 69, a: 0.7)

RGBA版本Swift 3/4

我喜欢卢卡的回答,因为我认为它是最优雅的。

然而,我不希望我的颜色指定在ARGB。我宁愿RGBA +,我也需要在处理字符串的情况下,为每个频道指定1个字符“#FFFA”。

这个版本还增加了错误抛出+剥离'#'字符如果它包含在字符串中。 这是我修改后的Swift表格。

public enum ColourParsingError: Error
{
    
    case invalidInput(String)
}
extension UIColor {
    public convenience init(hexString: String) throws
    {
        let hexString = hexString.replacingOccurrences(of: "#", with: "")
        let hex = hexString.trimmingCharacters(in:NSCharacterSet.alphanumerics.inverted)
        var int = UInt32()
        Scanner(string: hex).scanHexInt32(&int)
        let a, r, g, b: UInt32
        switch hex.count 
        {
        case 3: // RGB (12-bit)
            (r, g, b,a) = ((int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17,255)
        //iCSS specification in the form of #F0FA
        case 4: // RGB (24-bit)
            (r, g, b,a) = ((int >> 12) * 17, (int >> 8 & 0xF) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17)
        case 6: // RGB (24-bit)
            (r, g, b, a) = (int >> 16, int >> 8 & 0xFF, int & 0xFF,255)
        case 8: // ARGB (32-bit)
            (r, g, b, a) = (int >> 24, int >> 16 & 0xFF, int >> 8 & 0xFF, int & 0xFF)
        default:
            throw ColourParsingError.invalidInput("String is not a valid hex colour string: \(hexString)")
        }
        self.init(red: CGFloat(r) / 255, green: CGFloat(g) / 255, blue: CGFloat(b) / 255, alpha: CGFloat(a) / 255)
    }
}