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
这是UIColor的一个Swift扩展,它采用十六进制字符串:
import UIKit
extension UIColor {
convenience init(hexString: String) {
// Trim leading '#' if needed
var cleanedHexString = hexString
if hexString.hasPrefix("#") {
// cleanedHexString = dropFirst(hexString) // Swift 1.2
cleanedHexString = String(hexString.characters.dropFirst()) // Swift 2
}
// String -> UInt32
var rgbValue: UInt32 = 0
NSScanner(string: cleanedHexString).scanHexInt(&rgbValue)
// UInt32 -> R,G,B
let red = CGFloat((rgbValue >> 16) & 0xff) / 255.0
let green = CGFloat((rgbValue >> 08) & 0xff) / 255.0
let blue = CGFloat((rgbValue >> 00) & 0xff) / 255.0
self.init(red: red, green: green, blue: blue, alpha: 1.0)
}
}
其他回答
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)
)
}
斯威夫特5
extension UIColor{
/// Converting hex string to UIColor
///
/// - Parameter hexString: input hex string
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:
(a, r, g, b) = (255, (int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17)
case 6:
(a, r, g, b) = (255, int >> 16, int >> 8 & 0xFF, int & 0xFF)
case 8:
(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)
}
}
使用UIColor调用(hexString: "你的十六进制字符串")
Swift 5:你可以在Xcode中创建颜色,如下图所示:
您应该命名颜色,因为您通过名称引用了颜色。如图2所示:
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)
斯威夫特2.3: 用户界面颜色扩展。我认为这样更简单。
extension UIColor {
static func colorFromHex(hexString: String, alpha: CGFloat = 1) -> UIColor {
//checking if hex has 7 characters or not including '#'
if hexString.characters.count < 7 {
return UIColor.whiteColor()
}
//string by removing hash
let hexStringWithoutHash = hexString.substringFromIndex(hexString.startIndex.advancedBy(1))
//I am extracting three parts of hex color Red (first 2 characters), Green (middle 2 characters), Blue (last two characters)
let eachColor = [
hexStringWithoutHash.substringWithRange(hexStringWithoutHash.startIndex...hexStringWithoutHash.startIndex.advancedBy(1)),
hexStringWithoutHash.substringWithRange(hexStringWithoutHash.startIndex.advancedBy(2)...hexStringWithoutHash.startIndex.advancedBy(3)),
hexStringWithoutHash.substringWithRange(hexStringWithoutHash.startIndex.advancedBy(4)...hexStringWithoutHash.startIndex.advancedBy(5))]
let hexForEach = eachColor.map {CGFloat(Int($0, radix: 16) ?? 0)} //radix is base of numeric system you want to convert to, Hexadecimal has base 16
//return the color by making color
return UIColor(red: hexForEach[0] / 255, green: hexForEach[1] / 255, blue: hexForEach[2] / 255, alpha: alpha)
}
}
用法:
let color = UIColor.colorFromHex("#25ac09")