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
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)
你可以在UIColor上使用这个扩展,它将你的字符串(十六进制,RGBA)转换为UIColor,反之亦然。
extension UIColor {
//Convert RGBA String to UIColor object
//"rgbaString" must be separated by space "0.5 0.6 0.7 1.0" 50% of Red 60% of Green 70% of Blue Alpha 100%
public convenience init?(rgbaString : String){
self.init(ciColor: CIColor(string: rgbaString))
}
//Convert UIColor to RGBA String
func toRGBAString()-> String {
var r: CGFloat = 0
var g: CGFloat = 0
var b: CGFloat = 0
var a: CGFloat = 0
self.getRed(&r, green: &g, blue: &b, alpha: &a)
return "\(r) \(g) \(b) \(a)"
}
//return UIColor from Hexadecimal Color string
public convenience init?(hexString: String) {
let r, g, b, a: CGFloat
if hexString.hasPrefix("#") {
let start = hexString.index(hexString.startIndex, offsetBy: 1)
let hexColor = hexString.substring(from: start)
if hexColor.characters.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
}
// Convert UIColor to Hexadecimal String
func toHexString() -> String {
var r: CGFloat = 0
var g: CGFloat = 0
var b: CGFloat = 0
var a: CGFloat = 0
self.getRed(&r, green: &g, blue: &b, alpha: &a)
return String(
format: "%02X%02X%02X",
Int(r * 0xff),
Int(g * 0xff),
Int(b * 0xff))
}
}
Swift 4:结合Sulthan和Luca Torella的回答:
extension UIColor {
convenience init(hexFromString:String, alpha:CGFloat = 1.0) {
var cString:String = hexFromString.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
var rgbValue:UInt32 = 10066329 //color #999999 if string has wrong format
if (cString.hasPrefix("#")) {
cString.remove(at: cString.startIndex)
}
if ((cString.count) == 6) {
Scanner(string: cString).scanHexInt32(&rgbValue)
}
self.init(
red: CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0,
green: CGFloat((rgbValue & 0x00FF00) >> 8) / 255.0,
blue: CGFloat(rgbValue & 0x0000FF) / 255.0,
alpha: alpha
)
}
}
使用例子:
let myColor = UIColor(hexFromString: "4F9BF5")
let myColor = UIColor(hexFromString: "#4F9BF5")
let myColor = UIColor(hexFromString: "#4F9BF5", alpha: 0.5)
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
用户界面颜色:
extension UIColor {
convenience init(hex: Int) {
let components = (
R: CGFloat((hex >> 16) & 0xff) / 255,
G: CGFloat((hex >> 08) & 0xff) / 255,
B: CGFloat((hex >> 00) & 0xff) / 255
)
self.init(red: components.R, green: components.G, blue: components.B, alpha: 1)
}
}
CGColor:
extension CGColor {
class func colorWithHex(hex: Int) -> CGColorRef {
return UIColor(hex: hex).CGColor
}
}
使用
let purple = UIColor(hex: 0xAB47BC)