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

只是对第一个答案的一些补充

(还没有检查alpha,可能需要添加一个if next > 0xffffff):

extension UIColor {

struct COLORS_HEX {
    static let Primary = 0xffffff
    static let PrimaryDark = 0x000000
    static let Accent = 0xe89549
    static let AccentDark = 0xe27b2a
    static let TextWhiteSemiTransparent = 0x80ffffff
}

convenience init(red: Int, green: Int, blue: Int, alphaH: Int) {
    assert(red >= 0 && red <= 255, "Invalid red component")
    assert(green >= 0 && green <= 255, "Invalid green component")
    assert(blue >= 0 && blue <= 255, "Invalid blue component")
    assert(alphaH >= 0 && alphaH <= 255, "Invalid alpha component")

    self.init(red: CGFloat(red) / 255.0, green: CGFloat(green) / 255.0, blue: CGFloat(blue) / 255.0, alpha: CGFloat(alphaH) / 255.0)
}

convenience init(netHex:Int) {
    self.init(red:(netHex >> 16) & 0xff, green:(netHex >> 8) & 0xff, blue:netHex & 0xff, alphaH: (netHex >> 24) & 0xff)
}

}

其他回答

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)

#ffffff实际上是16进制表示法的3个颜色组件——红色ff,绿色ff和蓝色ff。你可以在Swift中使用0x前缀编写十六进制符号,例如0xFF

为了简化转换,让我们创建一个初始化式,它接受整数(0 - 255)值:

extension UIColor {
   convenience init(red: Int, green: Int, blue: Int) {
       assert(red >= 0 && red <= 255, "Invalid red component")
       assert(green >= 0 && green <= 255, "Invalid green component")
       assert(blue >= 0 && blue <= 255, "Invalid blue component")

       self.init(red: CGFloat(red) / 255.0, green: CGFloat(green) / 255.0, blue: CGFloat(blue) / 255.0, alpha: 1.0)
   }

   convenience init(rgb: Int) {
       self.init(
           red: (rgb >> 16) & 0xFF,
           green: (rgb >> 8) & 0xFF,
           blue: rgb & 0xFF
       )
   }
}

用法:

let color = UIColor(red: 0xFF, green: 0xFF, blue: 0xFF)
let color2 = UIColor(rgb: 0xFFFFFF)

如何得到alpha?

根据您的用例,您可以简单地使用本机UIColor。withAlphaComponent方法,例如:

let semitransparentBlack = UIColor(rgb: 0x000000).withAlphaComponent(0.5)

或者你可以在上面的方法中添加一个额外的(可选的)参数:

convenience init(red: Int, green: Int, blue: Int, a: CGFloat = 1.0) {
    self.init(
        red: CGFloat(red) / 255.0,
        green: CGFloat(green) / 255.0,
        blue: CGFloat(blue) / 255.0,
        alpha: a
    )
}

convenience init(rgb: Int, a: CGFloat = 1.0) {
    self.init(
        red: (rgb >> 16) & 0xFF,
        green: (rgb >> 8) & 0xFF,
        blue: rgb & 0xFF,
        a: a
    )
}

(我们不能将参数命名为alpha,因为与现有的初始化式名称冲突)。

称为:

let color = UIColor(red: 0xFF, green: 0xFF, blue: 0xFF, a: 0.5)
let color2 = UIColor(rgb: 0xFFFFFF, a: 0.5)

为了得到0-255的整数,我们可以

convenience init(red: Int, green: Int, blue: Int, a: Int = 0xFF) {
    self.init(
        red: CGFloat(red) / 255.0,
        green: CGFloat(green) / 255.0,
        blue: CGFloat(blue) / 255.0,
        alpha: CGFloat(a) / 255.0
    )
}

// let's suppose alpha is the first component (ARGB)
convenience init(argb: Int) {
    self.init(
        red: (argb >> 16) & 0xFF,
        green: (argb >> 8) & 0xFF,
        blue: argb & 0xFF,
        a: (argb >> 24) & 0xFF
    )
}

称为

let color = UIColor(red: 0xFF, green: 0xFF, blue: 0xFF, a: 0xFF)
let color2 = UIColor(argb: 0xFFFFFFFF)

或者是前面几种方法的组合。绝对没有必要使用字符串。

斯威夫特2.0

下面的代码是在xcode 7.2上测试的

import UIKit
extension UIColor{

    public convenience init?(colorCodeInHex: String, alpha: Float = 1.0){

        var filterColorCode:String =  colorCodeInHex.stringByReplacingOccurrencesOfString("#", withString: "")

        if  filterColorCode.characters.count != 6 {
            self.init(red: 0.0, green: 0.0, blue: 0.0, alpha: CGFloat(alpha))
            return
        }

        filterColorCode = filterColorCode.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString

        var range = Range(start: filterColorCode.startIndex.advancedBy(0), end: filterColorCode.startIndex.advancedBy(2))
        let rString = filterColorCode.substringWithRange(range)

        range = Range(start: filterColorCode.startIndex.advancedBy(2), end: filterColorCode.startIndex.advancedBy(4))
        let gString = filterColorCode.substringWithRange(range)


        range = Range(start: filterColorCode.startIndex.advancedBy(4), end: filterColorCode.startIndex.advancedBy(6))
        let bString = filterColorCode.substringWithRange(range)

        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)


        self.init(red: CGFloat(r) / 255.0, green: CGFloat(g) / 255.0, blue: CGFloat(b) / 255.0, alpha: CGFloat(alpha))
        return
    }
}

我做了一个小函数,把它放在我可以全局使用它的地方,在swift 2.1中工作得很好:

func getColorFromHex(rgbValue:UInt32)->UIColor{
   let red = CGFloat((rgbValue & 0xFF0000) >> 16)/255.0
   let green = CGFloat((rgbValue & 0xFF00) >> 8)/255.0
   let blue = CGFloat(rgbValue & 0xFF)/255.0

   return UIColor(red:red, green:green, blue:blue, alpha:1.0)
}

用法:

getColorFromHex(0xffffff)

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