我如何从十六进制字符串格式创建一个UIColor,如#00FF00?
当前回答
extension UIColor {
convenience init(hexaString: String, alpha: CGFloat = 1) {
let chars = Array(hexaString.dropFirst())
self.init(red: .init(strtoul(String(chars[0...1]),nil,16))/255,
green: .init(strtoul(String(chars[2...3]),nil,16))/255,
blue: .init(strtoul(String(chars[4...5]),nil,16))/255,
alpha: alpha)}
}
用法:
let redColor = UIColor(hexaString: "#FF0000") // r 1,0 g 0,0 b 0,0 a 1,0
let transparentRed = UIColor(hexaString: "#FF0000", alpha: 0.5) // r 1,0 g 0,0 b 0,0 a 0,5
另一种选择是将六值转换为无符号整数,并从中提取相应的值:
extension UIColor {
convenience init(hexaString: String, alpha: CGFloat = 1) {
self.init(hexa: UInt(hexaString.dropFirst(), radix: 16) ?? 0, alpha: alpha)
}
convenience init(hexa: UInt, alpha: CGFloat = 1) {
self.init(red: .init((hexa & 0xff0000) >> 16) / 255,
green: .init((hexa & 0xff00 ) >> 8) / 255,
blue: .init( hexa & 0xff ) / 255,
alpha: alpha)
}
}
let purpleColor = UIColor(hexaString: "#FF00FF") // r 1,0 g 0,0 b 1,0 a 1,0
let transparentYellow = UIColor(hexaString: "#FFFF00", alpha: 0.5) // r 1,0 g 1,0 b 0,0 a 0,5
其他回答
另一个带有alpha的版本
#define UIColorFromRGBA(rgbValue) [UIColor colorWithRed:((float)((rgbValue & 0xFF000000) >> 24))/255.0 green:((float)((rgbValue & 0xFF0000) >> 16))/255.0 blue:((float)((rgbValue & 0xFF00) >> 8 ))/255.0 alpha:((float)((rgbValue & 0xFF))/255.0)]
这是另一种选择。
- (UIColor *)colorWithRGBHex:(UInt32)hex
{
int r = (hex >> 16) & 0xFF;
int g = (hex >> 8) & 0xFF;
int b = (hex) & 0xFF;
return [UIColor colorWithRed:r / 255.0f
green:g / 255.0f
blue:b / 255.0f
alpha:1.0f];
}
抛光扩展从原来的答案@Tom 请随意在这里更新代码
extension UIColor{
convenience init (hexString:String) {
var cleanString:String = hexString.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString
if (cleanString.hasPrefix("#")) {
cleanString = cleanString.substringFromIndex(cleanString.startIndex.advancedBy(1))
}
if (cleanString.characters.count != 6) {
self.init()
}
else{
var rgbValue = UInt32()
let scanner = NSScanner(string: cleanString)
scanner.scanHexInt(&rgbValue)
self.init(
red: CGFloat((rgbValue & 0xFF0000) >> 16)/255.0,
green: CGFloat((rgbValue & 0xFF00) >> 8)/255.0,
blue: CGFloat(rgbValue & 0xFF)/255.0,
alpha: 1.0)
}
}
}
你可以做这样的扩展
extension UIColor {
convenience init(hex: UInt, alpha: CGFloat = 1) {
self.init(
red: CGFloat((hex & 0xFF0000) >> 16) / 255.0,
green: CGFloat((hex & 0x00FF00) >> 8) / 255.0,
blue: CGFloat(hex & 0x0000FF) / 255.0,
alpha: alpha
)
}
}
像这样用在任何地方
let color1 = UIColor(hex: 0xffffff)
let color2 = UIColor(hex: 0xffffff, alpha: 0.2)
Swift等价于@Tom的答案,尽管接收RGBA Int值以支持透明度:
func colorWithHex(aHex: UInt) -> UIColor
{
return UIColor(red: CGFloat((aHex & 0xFF000000) >> 24) / 255,
green: CGFloat((aHex & 0x00FF0000) >> 16) / 255,
blue: CGFloat((aHex & 0x0000FF00) >> 8) / 255,
alpha: CGFloat((aHex & 0x000000FF) >> 0) / 255)
}
//usage
var color = colorWithHex(0x7F00FFFF)
如果你想从string中使用它,你可以使用strtoul:
var hexString = "0x7F00FFFF"
let num = strtoul(hexString, nil, 16)
var colorFromString = colorWithHex(num)
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