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
我从这个回答中总结了一些想法,并针对iOS 13和Swift 5进行了更新。
extension UIColor {
convenience init(_ hex: String, alpha: CGFloat = 1.0) {
var cString = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
if cString.hasPrefix("#") { cString.removeFirst() }
if cString.count != 6 {
self.init("ff0000") // return red color for wrong hex input
return
}
var rgbValue: UInt64 = 0
Scanner(string: cString).scanHexInt64(&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)
}
}
然后你可以这样使用它:
UIColor("#ff0000") // with #
UIColor("ff0000") // without #
UIColor("ff0000", alpha: 0.5) // using optional alpha value
其他回答
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")
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)
)
}
我从这个回答中总结了一些想法,并针对iOS 13和Swift 5进行了更新。
extension UIColor {
convenience init(_ hex: String, alpha: CGFloat = 1.0) {
var cString = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
if cString.hasPrefix("#") { cString.removeFirst() }
if cString.count != 6 {
self.init("ff0000") // return red color for wrong hex input
return
}
var rgbValue: UInt64 = 0
Scanner(string: cString).scanHexInt64(&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)
}
}
然后你可以这样使用它:
UIColor("#ff0000") // with #
UIColor("ff0000") // without #
UIColor("ff0000", alpha: 0.5) // using optional alpha value
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)