我正在制作一个绘图应用程序，我想通过使用枚举来引用我的颜色。例如，每次我想要那种红色时，使用Colors.RedColor而不是输入值会更干净，更方便。但是，Swift的原始值枚举似乎不接受UIColor作为一种类型。有没有办法用枚举或类似的方法做到这一点？
答案 0 :(得分：44)
我是这样做的（基本上使用结构作为命名空间）：
extension UIColor {
struct MyTheme {
static var firstColor: UIColor { return UIColor(red: 1, green: 0, blue: 0, alpha: 1) }
static var secondColor: UIColor { return UIColor(red: 0, green: 1, blue: 0, alpha: 1) }
}
}
你用它就像：
UIColor.MyTheme.firstColor
因此，您可以在自定义主题中使用红色。
答案 1 :(得分：28)
如果您的颜色不是UIColor方便方法定义的颜色之一，则可以向UIColor添加扩展名：
extension UIColor {
static var firstColor: UIColor { return UIColor(red: 1, green: 0, blue: 0, alpha: 1) }
static var secondColor: UIColor { return UIColor(red: 0, green: 1, blue: 0, alpha: 1) }
}
// Usage
let myColor = UIColor.firstColor
答案 2 :(得分：15)
我使用计算属性来解决这个问题，这是我的代码
enum MyColor {
case navigationBarBackgroundColor
case navigationTintCololr
}
extension MyColor {
var value: UIColor {
get {
switch self {
case .navigationBarBackgroundColor:
return UIColor(red: 67/255, green: 173/255, blue: 247/255, alpha: 1.0)
case .navigationTintCololr:
return UIColor.white
}
}
}
}
然后我可以像这样使用MyColor：
MyColor.navigationBarBackgroundColor.value
答案 3 :(得分：5)
如何使用UIColor值制作Swift枚举？
这就是你如何使用UIColor值创建一个枚举：
import UIKit
final class Color: UIColor, RawRepresentable, ExpressibleByStringLiteral
{
// MARK:- ExpressibleByStringLiteral
typealias StringLiteralType = String
convenience init(stringLiteral: String) {
guard let (a,r,g,b) = Color.argb(hexColor: stringLiteral) else {
assertionFailure("Invalid string")
self.init(red: 0, green: 0, blue: 0, alpha: 0)
return
}
self.init(red: r, green: g, blue: b, alpha: a)
}
// MARK:- RawRepresentable
public typealias RawValue = String
convenience init?(rawValue: RawValue) {
guard let (a,r,g,b) = Color.argb(hexColor: rawValue) else { return nil }
self.init(red: r, green: g, blue: b, alpha: a)
}
var rawValue: RawValue {
return hexString()
}
// MARK:- Private
/// Return color components in range [0,1] for hexadecimal color strings.
/// - hexColor: case-insensitive string with format RGB, RRGGBB, or AARRGGBB.
private static func argb(hexColor: String) -> (CGFloat,CGFloat,CGFloat,CGFloat)?
{
let hexAlphabet = "0123456789abcdefABCDEF"
let hex = hexColor.trimmingCharacters(in: CharacterSet(charactersIn: hexAlphabet).inverted)
var int = UInt32()
Scanner(string: hex).scanHexInt32(&int)
let a, r, g, b: UInt32
switch hex.count {
case 3: (a, r, g, b) = (255, (int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17) // RGB
case 6: (a, r, g, b) = (255, int >> 16, int >> 8 & 0xFF, int & 0xFF) // RRGGBB
case 8: (a, r, g, b) = (int >> 24, int >> 16 & 0xFF, int >> 8 & 0xFF, int & 0xFF) // AARRGGBB
default: return nil
}
return (CGFloat(a)/255, CGFloat(r)/255, CGFloat(g)/255, CGFloat(b)/255)
}
private func hexString() -> String {
var red: CGFloat = 0
var green: CGFloat = 0
var blue: CGFloat = 0
var alpha: CGFloat = 0
if self.getRed(&red, green: &green, blue: &blue, alpha: &alpha) {
return String(format: "#%02X%02X%02X%02X", UInt8(red * 255), UInt8(green * 255), UInt8(blue * 255), UInt8(alpha * 255))
}
assertionFailure("Invalid colour space.")
return "#F00"
}
}
enum Colors: Color {
case red = "#F00"
// case blue = "#F00" // Raw value for enum case is not unique
}
let color3 = Color(rawValue: "#000") // RGB
let color6 = Color(rawValue: "#123456") // RRGGBB
let color8 = Color(rawValue: "#12345678") // AARRGGBB
print(Colors(rawValue:"#F00") as Any) // red
print(Colors(rawValue:"#FF0000") as Any) // red
print(Colors(rawValue:"#FFFF0000") as Any) // red
print(Colors(rawValue:"#ABC") as Any) // nil because it’s not a member of the enumeration
// print(Colors(rawValue:"#XYZ") as Any) // assertion on debug, black on release
print(Colors.red) // red
print(Colors.red.rawValue) // UIExtendedSRGBColorSpace 1 0 0 1
在
的帮助下答案 4 :(得分：1)
实际上我使用这样的实现，因为有两个原因对我来说非常方便，第一个我可以使用dex值而另一个是常量中的所有颜色
import UIKit
struct ColorPalette {
struct Gray {
static let Light = UIColor(netHex: 0x595959)
static let Medium = UIColor(netHex: 0x262626)
}
}
extension UIColor {
convenience init(red: Int, green: Int, blue: 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")
self.init(red: CGFloat(red) / 255.0, green: CGFloat(green) / 255.0, blue: CGFloat(blue) / 255.0, alpha: 1.0)
}
convenience init(netHex: Int) {
self.init(red: (netHex >> 16) & 0xff, green: (netHex >> 8) & 0xff, blue: netHex & 0xff)
}
}
用法
let backgroundGreyColor = ColorPalette.Gray.Medium.cgColor
答案 5 :(得分：1)
根据@ Jano的回答，我使用Int作为文字类型进行了改进：
import UIKit
public final class Colors: UIColor {
}
extension Colors: ExpressibleByIntegerLiteral {
public typealias IntegerLiteralType = Int
public convenience init(integerLiteral value: Int) {
let red = CGFloat((value & 0xFF0000FF) >> 24) / 0xFF
let green = CGFloat((value & 0x00FF00FF) >> 16) / 0xFF
let blue = CGFloat((value & 0x0000FFFF) >> 8) / 0xFF
let alpha = CGFloat(value & 0x00FF00FF) / 0xFF
self.init(red: red, green: green, blue: blue, alpha: alpha)
}
}
extension Colors: RawRepresentable {
public typealias RawValue = Int
public var rawValue: RawValue {
return hex
}
public convenience init?(rawValue: RawValue) {
self.init(integerLiteral: rawValue)
}
}
fileprivate extension UIColor {
var hex: Int {
var fRed: CGFloat = 0
var fGreen: CGFloat = 0
var fBlue: CGFloat = 0
var fAlpha: CGFloat = 0
if self.getRed(&fRed, green: &fGreen, blue: &fBlue, alpha: &fAlpha) {
let red = Int(fRed * 255.0)
let green = Int(fGreen * 255.0)
let blue = Int(fBlue * 255.0)
let alpha = Int(fAlpha * 255.0)
let rgb = (alpha << 24) + (red << 16) + (green << 8) + blue
return rgb
} else {
return 0x000000
}
}
}
public enum MainPalette: Colors {
case red = 0xFF0000ff
case white = 0xFFFFFFFF
}
public enum FeatureXPalette: Colors {
case blue = 0x024F9Eff
// case bluish = 0x024F9Eff // <- Can't do
case red = 0xFF0000ff
}
优点是它不允许重复颜色（作为真正的枚举）并且我也支持alpha。
如您所见，您可以为不同的调色板/方案创建多个枚举。如果您希望视图能够使用任何调色板，您只需添加协议：
protocol Color {
var color: UIColor { get }
}
extension MainPalette: Color {
var color: UIColor {
return rawValue
}
}
extension FeatureXPalette: Color {
var color: UIColor {
return rawValue
}
}
这样你就可以拥有一个接受协议的函数：
func printColorEquality(color1: Color, color2: Color) {
print(color1.color == color2.color)
}
let red1: Color = MainPalette.red
let red2: Color = FeatureXPalette.red
printColorEquality(color1: red1, color2: red2)
我还想做的是为方便起见添加静态变量：
extension MainPalette {
public static var brightRed: UIColor {
return MainPalette.red.color
}
}
给你一个更干净的api：
view.backgroundColor = MainPalette.brightRed
可以改进命名：你必须选择是否需要一个不错的便利api或为你的枚举命名。
答案 6 :(得分：0)
如果您想要返回多个值，请使用以下代码......绝对是 为我工作....
enum GetDriverStatus : String {
case ClockIn = "Clock In"
case TripStart = "Trip Start"
case BeaconTouchPlant = "Beacon Touch Plant"
case PickUp = "Pick Up"
case BeaconTouchSite = "Beacon Touch Site"
case BeaconLeftSite = "Beacon Left Site"
case DropOff = "Drop Off"
case BreakIn = "Break In"
case BreakOut = "Break Out"
case TripEnd = "Trip End"
case DayEnd = "Day End"
//case ClockOut = "Clock Out"
//Get data from ID
static var allValues: [GetDriverStatus] {
return [
.ClockIn,
.TripStart,
.BeaconTouchPlant,
.PickUp,
.BeaconTouchSite,
.BeaconLeftSite,
.DropOff,
.BreakIn,
.BreakOut,
.TripEnd,
.DayEnd
]
}
//Get Color
var colorAndStatus: (UIColor,String) {
get {
switch self {
case .ClockIn,.TripStart: //Idle
return (UIColor(red: 248/255, green: 39/255, blue: 71/255, alpha: 1.0),"Idle") //dark pink-red
case .BeaconTouchPlant,.PickUp:
return (UIColor(red: 46/255, green: 180/255, blue: 42/255, alpha: 1.0),"Picking up") //Green
case .BeaconTouchSite:
return (UIColor(red: 252/255, green: 172/255, blue: 0/255, alpha: 1.0),"On site") //orange
case .DropOff,.BeaconLeftSite:
return (UIColor(red: 12/255, green: 90/255, blue: 255/255, alpha: 1.0),"Dropping off") //blue
case .BreakIn,.BreakOut:
return (UIColor(red: 151/255, green: 151/255, blue: 151/255, alpha: 1.0),"On break") //warm-grey-two
case .TripEnd:
return (UIColor.black,"Trip end")
case .DayEnd:
return (UIColor.black,"Done for the day")
}
}
}
}
如何使用此代码
将.allvalues["index of your option"]传递到{0}位置以及UIColor为1位
String value答案 7 :(得分：0)
这可以更加简洁（并且应该这样做）：
extension UIColor
{
static let myColor = UIColor(displayP3Red: 0.0, green: 0.7, blue: 0.0, alpha: 1.0)
}
（其他任何返回UIColor的方法同样适用，不需要displayP3Red）
用法：
let someColor: UIColor = .myColor
答案 8 :(得分：0)
这个答案可能晚了，但是对于其他人来说发现这个问题。 我对上述答案不满意，因为将颜色添加为UIColors扩展并不总是您想要的，因为：
这是我想出的解决方案：
enum PencilColor {
case lightRed
case darkPurple
var associatedColor: UIColor {
switch self {
case .lightRed: return UIColor(red: 67/255, green: 173/255, blue: 247/255, alpha: 1.0)
case .darkPurple: return UIColor(red: 67/255, green: 173/255, blue: 247/255, alpha: 1.0)
}
}
}