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
斯威夫特4.0
使用这种单行方法
override func viewDidLoad() {
super.viewDidLoad()
let color = UIColor(hexColor: "FF00A0")
self.view.backgroundColor = color
}
你必须创建新的类或使用任何控制器,你需要使用十六进制颜色。这个扩展类为您提供UIColor,将十六进制转换为RGB颜色。
extension UIColor {
convenience init(hexColor: String) {
let scannHex = Scanner(string: hexColor)
var rgbValue: UInt64 = 0
scannHex.scanLocation = 0
scannHex.scanHexInt64(&rgbValue)
let r = (rgbValue & 0xff0000) >> 16
let g = (rgbValue & 0xff00) >> 8
let b = rgbValue & 0xff
self.init(
red: CGFloat(r) / 0xff,
green: CGFloat(g) / 0xff,
blue: CGFloat(b) / 0xff, alpha: 1
)
}
}
其他回答
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)
Swift 5 (Swift 4, Swift 3) UIColor扩展:
extension UIColor {
convenience init(hexString: String) {
let hex = hexString.trimmingCharacters(in: CharacterSet.alphanumerics.inverted)
var int = UInt64()
Scanner(string: hex).scanHexInt64(&int)
let a, r, g, b: UInt64
switch hex.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)
}
}
用法:
let darkGrey = UIColor(hexString: "#757575")
斯威夫特2。x版本:
extension UIColor {
convenience init(hexString: String) {
let hex = hexString.stringByTrimmingCharactersInSet(NSCharacterSet.alphanumericCharacterSet().invertedSet)
var int = UInt32()
NSScanner(string: hex).scanHexInt(&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)
}
}
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
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)
public static func hexStringToUIColor (hex:String) -> UIColor {
var cString:String = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
if (cString.hasPrefix("#")) {
cString.remove(at: cString.startIndex)
}
if ((cString.characters.count) == 6) {
var rgbValue:UInt32 = 0
Scanner(string: cString).scanHexInt32(&rgbValue)
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)
)
}else if ((cString.characters.count) == 8) {
var rgbValue:UInt32 = 0
Scanner(string: cString).scanHexInt32(&rgbValue)
return UIColor(
red: CGFloat((rgbValue & 0x00FF0000) >> 16) / 255.0,
green: CGFloat((rgbValue & 0x0000FF00) >> 8) / 255.0,
blue: CGFloat(rgbValue & 0x000000FF) / 255.0,
alpha: CGFloat((rgbValue & 0xFF000000) >> 24) / 255.0
)
}else{
return UIColor.gray
}
}
如何使用
var color: UIColor = hexStringToUIColor(hex: "#00ff00"); // Without transparency
var colorWithTransparency: UIColor = hexStringToUIColor(hex: "#dd00ff00"); // With transparency