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
extension UIColor {
convenience init(hex: Int, alpha: Double = 1.0) {
self.init(red: CGFloat((hex>>16)&0xFF)/255.0, green:CGFloat((hex>>8)&0xFF)/255.0, blue: CGFloat((hex)&0xFF)/255.0, alpha: CGFloat(255 * alpha) / 255)
}
}
使用这个扩展像:
let selectedColor = UIColor(hex: 0xFFFFFF)
let selectedColor = UIColor(hex: 0xFFFFFF, alpha: 0.5)
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")
extension UIColor {
public convenience init?(hex: String) {
let r, g, b, a: CGFloat
if hex.hasPrefix("#") {
let start = hex.index(hex.startIndex, offsetBy: 1)
let hexColor = String(hex[start...])
if hexColor.count == 8 {
let scanner = Scanner(string: hexColor)
var hexNumber: UInt64 = 0
if scanner.scanHexInt64(&hexNumber) {
r = CGFloat((hexNumber & 0xff000000) >> 24) / 255
g = CGFloat((hexNumber & 0x00ff0000) >> 16) / 255
b = CGFloat((hexNumber & 0x0000ff00) >> 8) / 255
a = CGFloat(hexNumber & 0x000000ff) / 255
self.init(red: r, green: g, blue: b, alpha: a)
return
}
}
}
return nil
}
}
用法:
let white = UIColor(hex: "#ffffff")
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)
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))
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