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
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)
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 {
public convenience init?(hex: String) {
let r, g, b, a: CGFloat
if hex.hasPrefix("#") {
let start = hex.index(hex.startIndex, offsetBy: 1)
let hexColor = String(hex[start...])
if hexColor.count == 8 {
let scanner = Scanner(string: hexColor)
var hexNumber: UInt64 = 0
if scanner.scanHexInt64(&hexNumber) {
r = CGFloat((hexNumber & 0xff000000) >> 24) / 255
g = CGFloat((hexNumber & 0x00ff0000) >> 16) / 255
b = CGFloat((hexNumber & 0x0000ff00) >> 8) / 255
a = CGFloat(hexNumber & 0x000000ff) / 255
self.init(red: r, green: g, blue: b, alpha: a)
return
}
}
}
return nil
}
}
用法:
let white = UIColor(hex: "#ffffff")
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
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)
)
}
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