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

斯威夫特5.0

你不能在Swift中直接使用#ffffff语法。以下是我用于网络相关项目的代码。支持alpha和三位数字。

用法示例(大写值也可以):

    let hex = "#FADE2B"  // yellow
    let color = NSColor(fromHex: hex)

支持的字符串格式:

"fff" // RGB "#fff" // #RGB "ffff" // RGBA "#ffff" // #RGBA . "ffffff" // RRGGBB . "#ffffff" // #RRGGBB . "ffffffff" // RRGGBBAA . "#ffffffff" // #RRGGBBAA .

数字代表红色,绿色,蓝色和阿尔法(像透明度)。对于iOS,用UIColor替换NSColor。

代码:


    extension NSColor {
        /// Initialises NSColor from a hexadecimal string. Color is clear if string is invalid.
        /// - Parameter fromHex: supported formats are "#RGB", "#RGBA", "#RRGGBB", "#RRGGBBAA", with or without the # character
        public convenience init(fromHex:String) {
            var r = 0, g = 0, b = 0, a = 255
            let offset = fromHex.hasPrefix("#") ? 1 : 0
            let ch = fromHex.map{$0}
            switch(ch.count - offset) {
            case 8:
                a = 16 * (ch[offset+6].hexDigitValue ?? 0) + (ch[offset+7].hexDigitValue ?? 0)
                fallthrough
            case 6:
                r = 16 * (ch[offset+0].hexDigitValue ?? 0) + (ch[offset+1].hexDigitValue ?? 0)
                g = 16 * (ch[offset+2].hexDigitValue ?? 0) + (ch[offset+3].hexDigitValue ?? 0)
                b = 16 * (ch[offset+4].hexDigitValue ?? 0) + (ch[offset+5].hexDigitValue ?? 0)
                break
            case 4:
                a = 16 * (ch[offset+3].hexDigitValue ?? 0) + (ch[offset+3].hexDigitValue ?? 0)
                fallthrough
            case 3:  // Three digit #0D3 is the same as six digit #00DD33
                r = 16 * (ch[offset+0].hexDigitValue ?? 0) + (ch[offset+0].hexDigitValue ?? 0)
                g = 16 * (ch[offset+1].hexDigitValue ?? 0) + (ch[offset+1].hexDigitValue ?? 0)
                b = 16 * (ch[offset+2].hexDigitValue ?? 0) + (ch[offset+2].hexDigitValue ?? 0)
                break
            default:
                a = 0
                break
            }
            self.init(red: CGFloat(r)/255, green: CGFloat(g)/255, blue: CGFloat(b)/255, alpha: CGFloat(a)/255)
            
        }
    }
    // Author: Andrew Kingdom

授权:CC BY

我发现这比下面的复制/粘贴更整洁

选择:

您可以删除#并将其存储为32位无符号整数字面量,由0x前缀表示,即0xffffff。不过,你仍然需要代码将其转换为颜色。

如果你想要一种非编程的方式来获取颜色:打开一个颜色选择器对话框,切换到colour Sliders > RGB Sliders,并将值粘贴/输入到“Hex color #”框中。(不要粘贴#散列符号。)

其他回答

#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。

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]

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

简单的颜色扩展Swift 5/SwiftUI

例子:

let myColor = Color(hex:0xF2C94C)

代码:

import Foundation
import SwiftUI

extension UIColor {
    convenience init(hex: Int) {
        let components = (
            R: CGFloat((hex >> 16) & 0xff) / 255,
            G: CGFloat((hex >> 08) & 0xff) / 255,
            B: CGFloat((hex >> 00) & 0xff) / 255
        )
        self.init(red: components.R, green: components.G, blue: components.B, alpha: 1)
    }
}

extension Color {
    public init(hex: Int) {
        self.init(UIColor(hex: hex))
   }
}
extension UIColor {
    public convenience init?(hex: String) {
        let r, g, b, a: CGFloat

        if hex.hasPrefix("#") {
            let start = hex.index(hex.startIndex, offsetBy: 1)
            let hexColor = String(hex[start...])

            if hexColor.count == 8 {
                let scanner = Scanner(string: hexColor)
                var hexNumber: UInt64 = 0

                if scanner.scanHexInt64(&hexNumber) {
                    r = CGFloat((hexNumber & 0xff000000) >> 24) / 255
                    g = CGFloat((hexNumber & 0x00ff0000) >> 16) / 255
                    b = CGFloat((hexNumber & 0x0000ff00) >> 8) / 255
                    a = CGFloat(hexNumber & 0x000000ff) / 255

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

        return nil
    }
}

用法:

let white = UIColor(hex: "#ffffff")