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

斯威夫特4.0

使用这种单行方法

override func viewDidLoad() {
    super.viewDidLoad()

   let color = UIColor(hexColor: "FF00A0")
   self.view.backgroundColor = color

}

你必须创建新的类或使用任何控制器,你需要使用十六进制颜色。这个扩展类为您提供UIColor,将十六进制转换为RGB颜色。

extension UIColor {
convenience init(hexColor: String) {
    let scannHex = Scanner(string: hexColor)
    var rgbValue: UInt64 = 0
    scannHex.scanLocation = 0
    scannHex.scanHexInt64(&rgbValue)
    let r = (rgbValue & 0xff0000) >> 16
    let g = (rgbValue & 0xff00) >> 8
    let b = rgbValue & 0xff
    self.init(
        red: CGFloat(r) / 0xff,
        green: CGFloat(g) / 0xff,
        blue: CGFloat(b) / 0xff, alpha: 1
    )
  }
}

其他回答

在Swift 2.0和Xcode 7.0.1中,你可以创建这个函数:

    // Creates a UIColor from a Hex string.
    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))
    }

然后这样使用它:

let color1 = colorWithHexString("#1F437C")

Swift 4更新

func colorWithHexString (hex:String) -> UIColor {

    var cString = hex.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines).uppercased()

    if (cString.hasPrefix("#")) {
        cString = (cString as NSString).substring(from: 1)
    }

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

    let rString = (cString as NSString).substring(to: 2)
    let gString = ((cString as NSString).substring(from: 2) as NSString).substring(to: 2)
    let bString = ((cString as NSString).substring(from: 4) as NSString).substring(to: 2)

    var r:CUnsignedInt = 0, g:CUnsignedInt = 0, b:CUnsignedInt = 0;
    Scanner(string: rString).scanHexInt32(&r)
    Scanner(string: gString).scanHexInt32(&g)
    Scanner(string: bString).scanHexInt32(&b)


    return UIColor(red: CGFloat(r) / 255.0, green: CGFloat(g) / 255.0, blue: CGFloat(b) / 255.0, alpha: CGFloat(1))
}

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)
    )
}

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)
    }
}

Swift 5.3 & SwiftUI:通过UIColor支持十六进制和CSS颜色名称

依据代码

斯威夫特包

SwiftUI包

字符串的例子:

橙子、酸橙、番茄等。 Clear, Transparent, nil和空字符串产生[UIColor clearColor] 美国广播公司 abc7 # abc7 00飞行符 # 00飞行符 00 ffff77

操场上的输出:

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: