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2.0:
在viewDidLoad ()
var viewColor:UIColor
viewColor = UIColor()
let colorInt:UInt
colorInt = 0x000000
viewColor = UIColorFromRGB(colorInt)
self.View.backgroundColor=viewColor
func UIColorFromRGB(rgbValue: UInt) -> UIColor {
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)
)
}
其他回答
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]
斯威夫特2.0:
做一个UIColor的扩展。
extension UIColor {
convenience init(hexString:String) {
let hexString:NSString = hexString.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
let scanner = NSScanner(string: hexString as String)
if (hexString.hasPrefix("#")) {
scanner.scanLocation = 1
}
var color:UInt32 = 0
scanner.scanHexInt(&color)
let mask = 0x000000FF
let r = Int(color >> 16) & mask
let g = Int(color >> 8) & mask
let b = Int(color) & mask
let red = CGFloat(r) / 255.0
let green = CGFloat(g) / 255.0
let blue = CGFloat(b) / 255.0
self.init(red:red, green:green, blue:blue, alpha:1)
}
func toHexString() -> String {
var r:CGFloat = 0
var g:CGFloat = 0
var b:CGFloat = 0
var a:CGFloat = 0
getRed(&r, green: &g, blue: &b, alpha: &a)
let rgb:Int = (Int)(r*255)<<16 | (Int)(g*255)<<8 | (Int)(b*255)<<0
return NSString(format:"#%06x", rgb) as String
}
}
用法:
//Hex to Color
let countPartColor = UIColor(hexString: "E43038")
//Color to Hex
let colorHexString = UIColor(red: 228, green: 48, blue: 56, alpha: 1.0).toHexString()
斯威夫特4.0
使用这种单行方法
override func viewDidLoad() {
super.viewDidLoad()
let color = UIColor(hexColor: "FF00A0")
self.view.backgroundColor = color
}
你必须创建新的类或使用任何控制器,你需要使用十六进制颜色。这个扩展类为您提供UIColor,将十六进制转换为RGB颜色。
extension UIColor {
convenience init(hexColor: String) {
let scannHex = Scanner(string: hexColor)
var rgbValue: UInt64 = 0
scannHex.scanLocation = 0
scannHex.scanHexInt64(&rgbValue)
let r = (rgbValue & 0xff0000) >> 16
let g = (rgbValue & 0xff00) >> 8
let b = rgbValue & 0xff
self.init(
red: CGFloat(r) / 0xff,
green: CGFloat(g) / 0xff,
blue: CGFloat(b) / 0xff, alpha: 1
)
}
}
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)
}
}
用户界面颜色:
extension UIColor {
convenience init(hex: Int) {
let components = (
R: CGFloat((hex >> 16) & 0xff) / 255,
G: CGFloat((hex >> 08) & 0xff) / 255,
B: CGFloat((hex >> 00) & 0xff) / 255
)
self.init(red: components.R, green: components.G, blue: components.B, alpha: 1)
}
}
CGColor:
extension CGColor {
class func colorWithHex(hex: Int) -> CGColorRef {
return UIColor(hex: hex).CGColor
}
}
使用
let purple = UIColor(hex: 0xAB47BC)