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
斯威夫特5.0
你不能在Swift中直接使用#ffffff语法。以下是我用于网络相关项目的代码。支持alpha和三位数字。
用法示例(大写值也可以):
let hex = "#FADE2B" // yellow
let color = NSColor(fromHex: hex)
支持的字符串格式:
"fff" // RGB "#fff" // #RGB "ffff" // RGBA "#ffff" // #RGBA . "ffffff" // RRGGBB . "#ffffff" // #RRGGBB . "ffffffff" // RRGGBBAA . "#ffffffff" // #RRGGBBAA .
数字代表红色,绿色,蓝色和阿尔法(像透明度)。对于iOS,用UIColor替换NSColor。
代码:
extension NSColor {
/// Initialises NSColor from a hexadecimal string. Color is clear if string is invalid.
/// - Parameter fromHex: supported formats are "#RGB", "#RGBA", "#RRGGBB", "#RRGGBBAA", with or without the # character
public convenience init(fromHex:String) {
var r = 0, g = 0, b = 0, a = 255
let offset = fromHex.hasPrefix("#") ? 1 : 0
let ch = fromHex.map{$0}
switch(ch.count - offset) {
case 8:
a = 16 * (ch[offset+6].hexDigitValue ?? 0) + (ch[offset+7].hexDigitValue ?? 0)
fallthrough
case 6:
r = 16 * (ch[offset+0].hexDigitValue ?? 0) + (ch[offset+1].hexDigitValue ?? 0)
g = 16 * (ch[offset+2].hexDigitValue ?? 0) + (ch[offset+3].hexDigitValue ?? 0)
b = 16 * (ch[offset+4].hexDigitValue ?? 0) + (ch[offset+5].hexDigitValue ?? 0)
break
case 4:
a = 16 * (ch[offset+3].hexDigitValue ?? 0) + (ch[offset+3].hexDigitValue ?? 0)
fallthrough
case 3: // Three digit #0D3 is the same as six digit #00DD33
r = 16 * (ch[offset+0].hexDigitValue ?? 0) + (ch[offset+0].hexDigitValue ?? 0)
g = 16 * (ch[offset+1].hexDigitValue ?? 0) + (ch[offset+1].hexDigitValue ?? 0)
b = 16 * (ch[offset+2].hexDigitValue ?? 0) + (ch[offset+2].hexDigitValue ?? 0)
break
default:
a = 0
break
}
self.init(red: CGFloat(r)/255, green: CGFloat(g)/255, blue: CGFloat(b)/255, alpha: CGFloat(a)/255)
}
}
// Author: Andrew Kingdom
授权:CC BY
我发现这比下面的复制/粘贴更整洁
选择:
您可以删除#并将其存储为32位无符号整数字面量,由0x前缀表示,即0xffffff。不过,你仍然需要代码将其转换为颜色。
如果你想要一种非编程的方式来获取颜色:打开一个颜色选择器对话框,切换到colour Sliders > RGB Sliders,并将值粘贴/输入到“Hex color #”框中。(不要粘贴#散列符号。)
其他回答
你可以在swift 5中使用它
斯威夫特5
import UIKit
extension UIColor {
static func hexStringToUIColor (hex:String) -> UIColor {
var cString:String = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
if (cString.hasPrefix("#")) {
cString.remove(at: cString.startIndex)
}
if ((cString.count) != 6) {
return UIColor.gray
}
var rgbValue:UInt32 = 0
Scanner(string: cString).scanHexInt32(&rgbValue)
return UIColor(
red: CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0,
green: CGFloat((rgbValue & 0x00FF00) >> 8) / 255.0,
blue: CGFloat(rgbValue & 0x0000FF) / 255.0,
alpha: CGFloat(1.0)
)
}
}
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]
这个答案展示了如何在Obj-C中实现。这座桥是要用的
let rgbValue = 0xFFEEDD
let r = Float((rgbValue & 0xFF0000) >> 16)/255.0
let g = Float((rgbValue & 0xFF00) >> 8)/255.0
let b = Float((rgbValue & 0xFF))/255.0
self.backgroundColor = UIColor(red:r, green: g, blue: b, alpha: 1.0)
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
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)