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

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))
}

其他回答

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
    )
  }
}

Swift 4:结合Sulthan和Luca Torella的回答:

extension UIColor {
    convenience init(hexFromString:String, alpha:CGFloat = 1.0) {
        var cString:String = hexFromString.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
        var rgbValue:UInt32 = 10066329 //color #999999 if string has wrong format

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

        if ((cString.count) == 6) {
            Scanner(string: cString).scanHexInt32(&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
        )
    }
}

使用例子:

let myColor = UIColor(hexFromString: "4F9BF5")

let myColor = UIColor(hexFromString: "#4F9BF5")

let myColor = UIColor(hexFromString: "#4F9BF5", alpha: 0.5)

以编程方式添加颜色的最简单方法是使用ColorLiteral。

只要像例子中那样添加ColorLiteral属性,Xcode就会提示你一个你可以选择的颜色列表。这样做的好处是代码更少,添加HEX值或RGB。您还将从故事板中获得最近使用的颜色。

例子: self.view.backgroundColor = ColorLiteral

这是我用的。适用于6和8字符的颜色字符串,带或不带#符号。在发布时默认为黑色,在调试时用无效字符串初始化时崩溃。

extension UIColor {
    public convenience init(hex: String) {
        var r: CGFloat = 0
        var g: CGFloat = 0
        var b: CGFloat = 0
        var a: CGFloat = 1

        let hexColor = hex.replacingOccurrences(of: "#", with: "")
        let scanner = Scanner(string: hexColor)
        var hexNumber: UInt64 = 0
        var valid = false

        if scanner.scanHexInt64(&hexNumber) {
            if hexColor.count == 8 {
                r = CGFloat((hexNumber & 0xff000000) >> 24) / 255
                g = CGFloat((hexNumber & 0x00ff0000) >> 16) / 255
                b = CGFloat((hexNumber & 0x0000ff00) >> 8) / 255
                a = CGFloat(hexNumber & 0x000000ff) / 255
                valid = true
            }
            else if hexColor.count == 6 {
                r = CGFloat((hexNumber & 0xff0000) >> 16) / 255
                g = CGFloat((hexNumber & 0x00ff00) >> 8) / 255
                b = CGFloat(hexNumber & 0x0000ff) / 255
                valid = true
            }
        }

        #if DEBUG
            assert(valid, "UIColor initialized with invalid hex string")
        #endif

        self.init(red: r, green: g, blue: b, alpha: a)
    }
}

用法:

UIColor(hex: "#75CC83FF")
UIColor(hex: "75CC83FF")
UIColor(hex: "#75CC83")
UIColor(hex: "75CC83")

我从这个回答中总结了一些想法,并针对iOS 13和Swift 5进行了更新。

extension UIColor {
  
  convenience init(_ hex: String, alpha: CGFloat = 1.0) {
    var cString = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
    
    if cString.hasPrefix("#") { cString.removeFirst() }
    
    if cString.count != 6 {
      self.init("ff0000") // return red color for wrong hex input
      return
    }
    
    var rgbValue: UInt64 = 0
    Scanner(string: cString).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)
  }

}

然后你可以这样使用它:

UIColor("#ff0000") // with #
UIColor("ff0000")  // without #
UIColor("ff0000", alpha: 0.5) // using optional alpha value