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

这是UIColor的一个Swift扩展,它采用十六进制字符串:

import UIKit

extension UIColor {

    convenience init(hexString: String) {
        // Trim leading '#' if needed
        var cleanedHexString = hexString
        if hexString.hasPrefix("#") {
//            cleanedHexString = dropFirst(hexString) // Swift 1.2
            cleanedHexString = String(hexString.characters.dropFirst()) // Swift 2
        }

        // String -> UInt32
        var rgbValue: UInt32 = 0
        NSScanner(string: cleanedHexString).scanHexInt(&rgbValue)

        // UInt32 -> R,G,B
        let red = CGFloat((rgbValue >> 16) & 0xff) / 255.0
        let green = CGFloat((rgbValue >> 08) & 0xff) / 255.0
        let blue = CGFloat((rgbValue >> 00) & 0xff) / 255.0

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

}

其他回答

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)
    )
}

斯威夫特5

extension UIColor{

/// Converting hex string to UIColor
///
/// - Parameter hexString: input hex string
convenience init(hexString: String) {
    let hex = hexString.trimmingCharacters(in: CharacterSet.alphanumerics.inverted)
    var int = UInt64()
    Scanner(string: hex).scanHexInt64(&int)
    let a, r, g, b: UInt64
    switch hex.count {
    case 3:    
        (a, r, g, b) = (255, (int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17)
    case 6: 
        (a, r, g, b) = (255, int >> 16, int >> 8 & 0xFF, int & 0xFF)
    case 8: 
        (a, r, g, b) = (int >> 24, int >> 16 & 0xFF, int >> 8 & 0xFF, int & 0xFF)
    default:
        (a, r, g, b) = (255, 0, 0, 0)
    }
    self.init(red: CGFloat(r) / 255, green: CGFloat(g) / 255, blue: CGFloat(b) / 255, alpha: CGFloat(a) / 255)
}
}

使用UIColor调用(hexString: "你的十六进制字符串")

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:

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)

斯威夫特2.3: 用户界面颜色扩展。我认为这样更简单。

extension UIColor {
    static func colorFromHex(hexString: String, alpha: CGFloat = 1) -> UIColor {
        //checking if hex has 7 characters or not including '#'
        if hexString.characters.count < 7 {
            return UIColor.whiteColor()
        }
        //string by removing hash
        let hexStringWithoutHash = hexString.substringFromIndex(hexString.startIndex.advancedBy(1))

        //I am extracting three parts of hex color Red (first 2 characters), Green (middle 2 characters), Blue (last two characters)
        let eachColor = [
            hexStringWithoutHash.substringWithRange(hexStringWithoutHash.startIndex...hexStringWithoutHash.startIndex.advancedBy(1)),
            hexStringWithoutHash.substringWithRange(hexStringWithoutHash.startIndex.advancedBy(2)...hexStringWithoutHash.startIndex.advancedBy(3)),
            hexStringWithoutHash.substringWithRange(hexStringWithoutHash.startIndex.advancedBy(4)...hexStringWithoutHash.startIndex.advancedBy(5))]

        let hexForEach = eachColor.map {CGFloat(Int($0, radix: 16) ?? 0)} //radix is base of numeric system you want to convert to, Hexadecimal has base 16

        //return the color by making color
        return UIColor(red: hexForEach[0] / 255, green: hexForEach[1] / 255, blue: hexForEach[2] / 255, alpha: alpha)
    }
}

用法:

let color = UIColor.colorFromHex("#25ac09")