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?
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
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")
其他回答
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)
在Swift 2.0和Xcode 7.0.1中,你可以创建这个函数:
// Creates a UIColor from a Hex string.
func colorWithHexString (hex:String) -> UIColor {
var cString:String = hex.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString
if (cString.hasPrefix("#")) {
cString = (cString as NSString).substringFromIndex(1)
}
if (cString.characters.count != 6) {
return UIColor.grayColor()
}
let rString = (cString as NSString).substringToIndex(2)
let gString = ((cString as NSString).substringFromIndex(2) as NSString).substringToIndex(2)
let bString = ((cString as NSString).substringFromIndex(4) as NSString).substringToIndex(2)
var r:CUnsignedInt = 0, g:CUnsignedInt = 0, b:CUnsignedInt = 0;
NSScanner(string: rString).scanHexInt(&r)
NSScanner(string: gString).scanHexInt(&g)
NSScanner(string: bString).scanHexInt(&b)
return UIColor(red: CGFloat(r) / 255.0, green: CGFloat(g) / 255.0, blue: CGFloat(b) / 255.0, alpha: CGFloat(1))
}
然后这样使用它:
let color1 = colorWithHexString("#1F437C")
Swift 4更新
func colorWithHexString (hex:String) -> UIColor {
var cString = hex.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines).uppercased()
if (cString.hasPrefix("#")) {
cString = (cString as NSString).substring(from: 1)
}
if (cString.characters.count != 6) {
return UIColor.gray
}
let rString = (cString as NSString).substring(to: 2)
let gString = ((cString as NSString).substring(from: 2) as NSString).substring(to: 2)
let bString = ((cString as NSString).substring(from: 4) as NSString).substring(to: 2)
var r:CUnsignedInt = 0, g:CUnsignedInt = 0, b:CUnsignedInt = 0;
Scanner(string: rString).scanHexInt32(&r)
Scanner(string: gString).scanHexInt32(&g)
Scanner(string: bString).scanHexInt32(&b)
return UIColor(red: CGFloat(r) / 255.0, green: CGFloat(g) / 255.0, blue: CGFloat(b) / 255.0, alpha: CGFloat(1))
}
extension UIColor {
convenience init(r: CGFloat, g: CGFloat, b: CGFloat, a: CGFloat = 1) {
self.init(red: r/255, green: g/255, blue: b/255, alpha: a)
}
convenience init(hex: Int, alpha: CGFloat = 1) {
self.init(r: CGFloat((hex >> 16) & 0xff), g: CGFloat((hex >> 08) & 0xff), b: CGFloat((hex >> 00) & 0xff), a: alpha)
}
}
iOS 14, SwiftUI 2.0, swift 5.1, Xcode beta12
extension Color {
static func hexColour(hexValue:UInt32)->Color
{
let red = Double((hexValue & 0xFF0000) >> 16) / 255.0
let green = Double((hexValue & 0xFF00) >> 8) / 255.0
let blue = Double(hexValue & 0xFF) / 255.0
return Color(red:red, green:green, blue:blue)
}
}
用十六进制数表示
let red = Color.hexColour(hexValue: 0xFF0000)