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

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

其他回答

这个答案展示了如何在Obj-C中实现。这座桥是要用的

let rgbValue = 0xFFEEDD
let r = Float((rgbValue & 0xFF0000) >> 16)/255.0
let g = Float((rgbValue & 0xFF00) >> 8)/255.0
let b = Float((rgbValue & 0xFF))/255.0
self.backgroundColor = UIColor(red:r, green: g, blue: b, alpha: 1.0)

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)

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

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

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)
    }

}