带验证的十六进制

根据爱德华多的回答

细节

Xcode 10.0, Swift 4.2 Xcode 10.2.1 (10E1001)

解决方案

import UIKit

extension UIColor {

    convenience init(r: UInt8, g: UInt8, b: UInt8, alpha: CGFloat = 1.0) {
        let divider: CGFloat = 255.0
        self.init(red: CGFloat(r)/divider, green: CGFloat(g)/divider, blue: CGFloat(b)/divider, alpha: alpha)
    }

    private convenience init(rgbWithoutValidation value: Int32, alpha: CGFloat = 1.0) {
        self.init(
            r: UInt8((value & 0xFF0000) >> 16),
            g: UInt8((value & 0x00FF00) >> 8),
            b: UInt8(value & 0x0000FF),
            alpha: alpha
        )
    }

    convenience init?(rgb: Int32, alpha: CGFloat = 1.0) {
        if rgb > 0xFFFFFF || rgb < 0 { return nil }
        self.init(rgbWithoutValidation: rgb, alpha: alpha)
    }

    convenience init?(hex: String, alpha: CGFloat = 1.0) {
        var charSet = CharacterSet.whitespacesAndNewlines
        charSet.insert("#")
        let _hex = hex.trimmingCharacters(in: charSet)
        guard _hex.range(of: "^[0-9A-Fa-f]{6}$", options: .regularExpression) != nil else { return nil }
        var rgb: UInt32 = 0
        Scanner(string: _hex).scanHexInt32(&rgb)
        self.init(rgbWithoutValidation: Int32(rgb), alpha: alpha)
    }
}

使用

let alpha: CGFloat = 1.0

// Hex
print(UIColor(rgb: 0x4F9BF5) ?? "nil")
print(UIColor(rgb: 0x4F9BF5, alpha: alpha) ?? "nil")
print(UIColor(rgb: 5217269) ?? "nil")
print(UIColor(rgb: -5217269) ?? "nil")                  // = nil
print(UIColor(rgb: 0xFFFFFF1) ?? "nil")                 // = nil

// String
print(UIColor(hex: "4F9BF5") ?? "nil")
print(UIColor(hex: "4F9BF5", alpha: alpha) ?? "nil")
print(UIColor(hex: "#4F9BF5") ?? "nil")
print(UIColor(hex: "#4F9BF5", alpha: alpha) ?? "nil")
print(UIColor(hex: "#4F9BF56") ?? "nil")                // = nil
print(UIColor(hex: "#blabla") ?? "nil")                 // = nil

// RGB
print(UIColor(r: 79, g: 155, b: 245))
print(UIColor(r: 79, g: 155, b: 245, alpha: alpha))
//print(UIColor(r: 792, g: 155, b: 245, alpha: alpha))  // Compiler will throw an error, r,g,b = [0...255]

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)
    }
}

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

extension UIColor {

      convenience init(hex: Int, alpha: Double = 1.0) {

      self.init(red: CGFloat((hex>>16)&0xFF)/255.0, green:CGFloat((hex>>8)&0xFF)/255.0, blue: CGFloat((hex)&0xFF)/255.0, alpha:  CGFloat(255 * alpha) / 255)
     }
}

使用这个扩展像:

let selectedColor = UIColor(hex: 0xFFFFFF)
let selectedColor = UIColor(hex: 0xFFFFFF, alpha: 0.5)

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")

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)