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
UIColor扩展,这将大大帮助你!(4.0版本:斯威夫特)
import UIKit
extension UIColor {
/// rgb颜色
convenience init(r: CGFloat, g: CGFloat, b: CGFloat) {
self.init(red: r/255.0 ,green: g/255.0 ,blue: b/255.0 ,alpha:1.0)
}
/// 纯色(用于灰色)
convenience init(gray: CGFloat) {
self.init(red: gray/255.0 ,green: gray/255.0 ,blue: gray/255.0 ,alpha:1.0)
}
/// 随机色
class func randomCGColor() -> UIColor {
return UIColor(r: CGFloat(arc4random_uniform(256)), g: CGFloat(arc4random_uniform(256)), b: CGFloat(arc4random_uniform(256)))
}
/// hex颜色-Int
convenience init(hex:Int, alpha:CGFloat = 1.0) {
self.init(
red: CGFloat((hex & 0xFF0000) >> 16) / 255.0,
green: CGFloat((hex & 0x00FF00) >> 8) / 255.0,
blue: CGFloat((hex & 0x0000FF) >> 0) / 255.0,
alpha: alpha
)
}
/// hex颜色-String
convenience init(hexString: String){
var red: CGFloat = 0.0
var green: CGFloat = 0.0
var blue: CGFloat = 0.0
var alpha: CGFloat = 1.0
let scanner = Scanner(string: hexString)
var hexValue: CUnsignedLongLong = 0
if scanner.scanHexInt64(&hexValue) {
switch (hexString.characters.count) {
case 3:
red = CGFloat((hexValue & 0xF00) >> 8) / 15.0
green = CGFloat((hexValue & 0x0F0) >> 4) / 15.0
blue = CGFloat(hexValue & 0x00F) / 15.0
case 4:
red = CGFloat((hexValue & 0xF000) >> 12) / 15.0
green = CGFloat((hexValue & 0x0F00) >> 8) / 15.0
blue = CGFloat((hexValue & 0x00F0) >> 4) / 15.0
alpha = CGFloat(hexValue & 0x000F) / 15.0
case 6:
red = CGFloat((hexValue & 0xFF0000) >> 16) / 255.0
green = CGFloat((hexValue & 0x00FF00) >> 8) / 255.0
blue = CGFloat(hexValue & 0x0000FF) / 255.0
case 8:
alpha = CGFloat((hexValue & 0xFF000000) >> 24) / 255.0
red = CGFloat((hexValue & 0x00FF0000) >> 16) / 255.0
green = CGFloat((hexValue & 0x0000FF00) >> 8) / 255.0
blue = CGFloat(hexValue & 0x000000FF) / 255.0
default:
log.info("Invalid RGB string, number of characters after '#' should be either 3, 4, 6 or 8")
}
} else {
log.error("Scan hex error")
}
self.init(red:red, green:green, blue:blue, alpha:alpha)
}}
其他回答
斯威夫特2.0:
做一个UIColor的扩展。
extension UIColor {
convenience init(hexString:String) {
let hexString:NSString = hexString.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
let scanner = NSScanner(string: hexString as String)
if (hexString.hasPrefix("#")) {
scanner.scanLocation = 1
}
var color:UInt32 = 0
scanner.scanHexInt(&color)
let mask = 0x000000FF
let r = Int(color >> 16) & mask
let g = Int(color >> 8) & mask
let b = Int(color) & mask
let red = CGFloat(r) / 255.0
let green = CGFloat(g) / 255.0
let blue = CGFloat(b) / 255.0
self.init(red:red, green:green, blue:blue, alpha:1)
}
func toHexString() -> String {
var r:CGFloat = 0
var g:CGFloat = 0
var b:CGFloat = 0
var a:CGFloat = 0
getRed(&r, green: &g, blue: &b, alpha: &a)
let rgb:Int = (Int)(r*255)<<16 | (Int)(g*255)<<8 | (Int)(b*255)<<0
return NSString(format:"#%06x", rgb) as String
}
}
用法:
//Hex to Color
let countPartColor = UIColor(hexString: "E43038")
//Color to Hex
let colorHexString = UIColor(red: 228, green: 48, blue: 56, alpha: 1.0).toHexString()
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)
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)
RGBA版本Swift 3/4
我喜欢卢卡的回答,因为我认为它是最优雅的。
然而,我不希望我的颜色指定在ARGB。我宁愿RGBA +,我也需要在处理字符串的情况下,为每个频道指定1个字符“#FFFA”。
这个版本还增加了错误抛出+剥离'#'字符如果它包含在字符串中。 这是我修改后的Swift表格。
public enum ColourParsingError: Error
{
case invalidInput(String)
}
extension UIColor {
public convenience init(hexString: String) throws
{
let hexString = hexString.replacingOccurrences(of: "#", with: "")
let hex = hexString.trimmingCharacters(in:NSCharacterSet.alphanumerics.inverted)
var int = UInt32()
Scanner(string: hex).scanHexInt32(&int)
let a, r, g, b: UInt32
switch hex.count
{
case 3: // RGB (12-bit)
(r, g, b,a) = ((int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17,255)
//iCSS specification in the form of #F0FA
case 4: // RGB (24-bit)
(r, g, b,a) = ((int >> 12) * 17, (int >> 8 & 0xF) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17)
case 6: // RGB (24-bit)
(r, g, b, a) = (int >> 16, int >> 8 & 0xFF, int & 0xFF,255)
case 8: // ARGB (32-bit)
(r, g, b, a) = (int >> 24, int >> 16 & 0xFF, int >> 8 & 0xFF, int & 0xFF)
default:
throw ColourParsingError.invalidInput("String is not a valid hex colour string: \(hexString)")
}
self.init(red: CGFloat(r) / 255, green: CGFloat(g) / 255, blue: CGFloat(b) / 255, alpha: CGFloat(a) / 255)
}
}
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)