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
另一种方法
斯威夫特3.0
为UIColor写一个扩展
// To change the HexaDecimal value to Corresponding Color
extension UIColor
{
class func uicolorFromHex(_ rgbValue:UInt32, alpha : CGFloat)->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: alpha)
}
}
你可以像这样用hex直接创建UIColor
let carrot = UIColor.uicolorFromHex(0xe67e22, alpha: 1))
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]
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)
}
}
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)
}
}
#ffffff实际上是16进制表示法的3个颜色组件——红色ff,绿色ff和蓝色ff。你可以在Swift中使用0x前缀编写十六进制符号,例如0xFF
为了简化转换,让我们创建一个初始化式,它接受整数(0 - 255)值:
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(rgb: Int) {
self.init(
red: (rgb >> 16) & 0xFF,
green: (rgb >> 8) & 0xFF,
blue: rgb & 0xFF
)
}
}
用法:
let color = UIColor(red: 0xFF, green: 0xFF, blue: 0xFF)
let color2 = UIColor(rgb: 0xFFFFFF)
如何得到alpha?
根据您的用例,您可以简单地使用本机UIColor。withAlphaComponent方法,例如:
let semitransparentBlack = UIColor(rgb: 0x000000).withAlphaComponent(0.5)
或者你可以在上面的方法中添加一个额外的(可选的)参数:
convenience init(red: Int, green: Int, blue: Int, a: CGFloat = 1.0) {
self.init(
red: CGFloat(red) / 255.0,
green: CGFloat(green) / 255.0,
blue: CGFloat(blue) / 255.0,
alpha: a
)
}
convenience init(rgb: Int, a: CGFloat = 1.0) {
self.init(
red: (rgb >> 16) & 0xFF,
green: (rgb >> 8) & 0xFF,
blue: rgb & 0xFF,
a: a
)
}
(我们不能将参数命名为alpha,因为与现有的初始化式名称冲突)。
称为:
let color = UIColor(red: 0xFF, green: 0xFF, blue: 0xFF, a: 0.5)
let color2 = UIColor(rgb: 0xFFFFFF, a: 0.5)
为了得到0-255的整数,我们可以
convenience init(red: Int, green: Int, blue: Int, a: Int = 0xFF) {
self.init(
red: CGFloat(red) / 255.0,
green: CGFloat(green) / 255.0,
blue: CGFloat(blue) / 255.0,
alpha: CGFloat(a) / 255.0
)
}
// let's suppose alpha is the first component (ARGB)
convenience init(argb: Int) {
self.init(
red: (argb >> 16) & 0xFF,
green: (argb >> 8) & 0xFF,
blue: argb & 0xFF,
a: (argb >> 24) & 0xFF
)
}
称为
let color = UIColor(red: 0xFF, green: 0xFF, blue: 0xFF, a: 0xFF)
let color2 = UIColor(argb: 0xFFFFFFFF)
或者是前面几种方法的组合。绝对没有必要使用字符串。