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
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]
其他回答
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)
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]
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)
Swift 5 (Swift 4, Swift 3) UIColor扩展:
extension UIColor {
convenience init(hexString: String) {
let hex = hexString.trimmingCharacters(in: CharacterSet.alphanumerics.inverted)
var int = UInt64()
Scanner(string: hex).scanHexInt64(&int)
let a, r, g, b: UInt64
switch hex.count {
case 3: // RGB (12-bit)
(a, r, g, b) = (255, (int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17)
case 6: // RGB (24-bit)
(a, r, g, b) = (255, int >> 16, int >> 8 & 0xFF, int & 0xFF)
case 8: // ARGB (32-bit)
(a, r, g, b) = (int >> 24, int >> 16 & 0xFF, int >> 8 & 0xFF, int & 0xFF)
default:
(a, r, g, b) = (255, 0, 0, 0)
}
self.init(red: CGFloat(r) / 255, green: CGFloat(g) / 255, blue: CGFloat(b) / 255, alpha: CGFloat(a) / 255)
}
}
用法:
let darkGrey = UIColor(hexString: "#757575")
斯威夫特2。x版本:
extension UIColor {
convenience init(hexString: String) {
let hex = hexString.stringByTrimmingCharactersInSet(NSCharacterSet.alphanumericCharacterSet().invertedSet)
var int = UInt32()
NSScanner(string: hex).scanHexInt(&int)
let a, r, g, b: UInt32
switch hex.characters.count {
case 3: // RGB (12-bit)
(a, r, g, b) = (255, (int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17)
case 6: // RGB (24-bit)
(a, r, g, b) = (255, int >> 16, int >> 8 & 0xFF, int & 0xFF)
case 8: // ARGB (32-bit)
(a, r, g, b) = (int >> 24, int >> 16 & 0xFF, int >> 8 & 0xFF, int & 0xFF)
default:
(a, r, g, b) = (255, 0, 0, 0)
}
self.init(red: CGFloat(r) / 255, green: CGFloat(g) / 255, blue: CGFloat(b) / 255, alpha: CGFloat(a) / 255)
}
}
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)
}
}
推荐文章
- 如何删除默认的导航栏空间在SwiftUI导航视图
- 如何在iOS中使用Swift编程segue
- Swift -整数转换为小时/分钟/秒
- 如何舍入一个双到最近的Int在迅速?
- 扁平化数组的数组在Swift
- Swift:声明一个空字典
- 在成功提交我的应用程序后，“太多符号文件”
- 从数组中随机选择一个元素
- 首先添加一个UIView，甚至是导航栏
- 我如何改变UIButton标题颜色?
- 在Swift中如何调用GCD主线程上的参数方法?
- NSLayoutConstraints是可动画的吗?
- iOS -构建失败，CocoaPods无法找到头文件
- CFNetwork SSLHandshake iOS 9失败
- swift语言中的结构与类
