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
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)
}
}
其他回答
我从这个回答中总结了一些想法,并针对iOS 13和Swift 5进行了更新。
extension UIColor {
convenience init(_ hex: String, alpha: CGFloat = 1.0) {
var cString = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
if cString.hasPrefix("#") { cString.removeFirst() }
if cString.count != 6 {
self.init("ff0000") // return red color for wrong hex input
return
}
var rgbValue: UInt64 = 0
Scanner(string: cString).scanHexInt64(&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)
}
}
然后你可以这样使用它:
UIColor("#ff0000") // with #
UIColor("ff0000") // without #
UIColor("ff0000", alpha: 0.5) // using optional alpha value
带验证的十六进制
根据爱德华多的回答
细节
Xcode 10.0, Swift 4.2 Xcode 10.2.1 (10E1001)
解决方案
import UIKit
extension UIColor {
convenience init(r: UInt8, g: UInt8, b: UInt8, alpha: CGFloat = 1.0) {
let divider: CGFloat = 255.0
self.init(red: CGFloat(r)/divider, green: CGFloat(g)/divider, blue: CGFloat(b)/divider, alpha: alpha)
}
private convenience init(rgbWithoutValidation value: Int32, alpha: CGFloat = 1.0) {
self.init(
r: UInt8((value & 0xFF0000) >> 16),
g: UInt8((value & 0x00FF00) >> 8),
b: UInt8(value & 0x0000FF),
alpha: alpha
)
}
convenience init?(rgb: Int32, alpha: CGFloat = 1.0) {
if rgb > 0xFFFFFF || rgb < 0 { return nil }
self.init(rgbWithoutValidation: rgb, alpha: alpha)
}
convenience init?(hex: String, alpha: CGFloat = 1.0) {
var charSet = CharacterSet.whitespacesAndNewlines
charSet.insert("#")
let _hex = hex.trimmingCharacters(in: charSet)
guard _hex.range(of: "^[0-9A-Fa-f]{6}$", options: .regularExpression) != nil else { return nil }
var rgb: UInt32 = 0
Scanner(string: _hex).scanHexInt32(&rgb)
self.init(rgbWithoutValidation: Int32(rgb), alpha: alpha)
}
}
使用
let alpha: CGFloat = 1.0
// Hex
print(UIColor(rgb: 0x4F9BF5) ?? "nil")
print(UIColor(rgb: 0x4F9BF5, alpha: alpha) ?? "nil")
print(UIColor(rgb: 5217269) ?? "nil")
print(UIColor(rgb: -5217269) ?? "nil") // = nil
print(UIColor(rgb: 0xFFFFFF1) ?? "nil") // = nil
// String
print(UIColor(hex: "4F9BF5") ?? "nil")
print(UIColor(hex: "4F9BF5", alpha: alpha) ?? "nil")
print(UIColor(hex: "#4F9BF5") ?? "nil")
print(UIColor(hex: "#4F9BF5", alpha: alpha) ?? "nil")
print(UIColor(hex: "#4F9BF56") ?? "nil") // = nil
print(UIColor(hex: "#blabla") ?? "nil") // = nil
// RGB
print(UIColor(r: 79, g: 155, b: 245))
print(UIColor(r: 79, g: 155, b: 245, alpha: alpha))
//print(UIColor(r: 792, g: 155, b: 245, alpha: alpha)) // Compiler will throw an error, r,g,b = [0...255]
只是对第一个答案的一些补充
(还没有检查alpha,可能需要添加一个if next > 0xffffff):
extension UIColor {
struct COLORS_HEX {
static let Primary = 0xffffff
static let PrimaryDark = 0x000000
static let Accent = 0xe89549
static let AccentDark = 0xe27b2a
static let TextWhiteSemiTransparent = 0x80ffffff
}
convenience init(red: Int, green: Int, blue: Int, alphaH: Int) {
assert(red >= 0 && red <= 255, "Invalid red component")
assert(green >= 0 && green <= 255, "Invalid green component")
assert(blue >= 0 && blue <= 255, "Invalid blue component")
assert(alphaH >= 0 && alphaH <= 255, "Invalid alpha component")
self.init(red: CGFloat(red) / 255.0, green: CGFloat(green) / 255.0, blue: CGFloat(blue) / 255.0, alpha: CGFloat(alphaH) / 255.0)
}
convenience init(netHex:Int) {
self.init(red:(netHex >> 16) & 0xff, green:(netHex >> 8) & 0xff, blue:netHex & 0xff, alphaH: (netHex >> 24) & 0xff)
}
}
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)