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
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)
}
}
其他回答
你可以在swift 5中使用它
斯威夫特5
import UIKit
extension UIColor {
static func hexStringToUIColor (hex:String) -> UIColor {
var cString:String = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
if (cString.hasPrefix("#")) {
cString.remove(at: cString.startIndex)
}
if ((cString.count) != 6) {
return UIColor.gray
}
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)
)
}
}
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)
我做了一个小函数,把它放在我可以全局使用它的地方,在swift 2.1中工作得很好:
func getColorFromHex(rgbValue:UInt32)->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:1.0)
}
用法:
getColorFromHex(0xffffff)
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)
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)
}
}