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

另一种方法

斯威夫特3.0

为UIColor写一个扩展

// To change the HexaDecimal value to Corresponding Color
extension UIColor
{
    class func uicolorFromHex(_ rgbValue:UInt32, alpha : CGFloat)->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: alpha)
    }
}

你可以像这样用hex直接创建UIColor

let carrot = UIColor.uicolorFromHex(0xe67e22, alpha: 1))

其他回答

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)

最新swift3版本

        extension UIColor {
convenience init(hexString: String) {
    let hex = hexString.trimmingCharacters(in: CharacterSet.alphanumerics.inverted)
    var int = UInt32()
    Scanner(string: hex).scanHexInt32(&int)
    let a, r, g, b: UInt32
    switch hex.characters.count {
    case 3: // RGB (12-bit)
        (a, r, g, b) = (255, (int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17)
    case 6: // RGB (24-bit)
        (a, r, g, b) = (255, int >> 16, int >> 8 & 0xFF, int & 0xFF)
    case 8: // ARGB (32-bit)
        (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)
}
}

在你的类或任何你把hexcolor转换为uicolor的地方使用这种方法

             let color1 = UIColor(hexString: "#FF323232")

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)
    )
}

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)

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: