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

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)

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)

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)
    )
}

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)
    }

}

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

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)
    }
}