我如何从十六进制字符串格式创建一个UIColor,如#00FF00?
当前回答
斯威夫特4
你可以像这样在扩展中创建一个非常方便的构造函数:
extension UIColor {
convenience init(hexString: String, alpha: CGFloat = 1.0) {
var hexInt: UInt32 = 0
let scanner = Scanner(string: hexString)
scanner.charactersToBeSkipped = CharacterSet(charactersIn: "#")
scanner.scanHexInt32(&hexInt)
let red = CGFloat((hexInt & 0xff0000) >> 16) / 255.0
let green = CGFloat((hexInt & 0xff00) >> 8) / 255.0
let blue = CGFloat((hexInt & 0xff) >> 0) / 255.0
let alpha = alpha
self.init(red: red, green: green, blue: blue, alpha: alpha)
}
}
以后再用
let color = UIColor(hexString: "#AABBCCDD")
其他回答
一个简明的解决方案:
// Assumes input like "#00FF00" (#RRGGBB).
+ (UIColor *)colorFromHexString:(NSString *)hexString {
unsigned rgbValue = 0;
NSScanner *scanner = [NSScanner scannerWithString:hexString];
[scanner setScanLocation:1]; // bypass '#' character
[scanner scanHexInt:&rgbValue];
return [UIColor colorWithRed:((rgbValue & 0xFF0000) >> 16)/255.0 green:((rgbValue & 0xFF00) >> 8)/255.0 blue:(rgbValue & 0xFF)/255.0 alpha:1.0];
}
swift 2.0+。 这段代码对我来说很好。
extension UIColor {
/// UIColor(hexString: "#cc0000")
internal convenience init?(hexString:String) {
guard hexString.characters[hexString.startIndex] == Character("#") else {
return nil
}
guard hexString.characters.count == "#000000".characters.count else {
return nil
}
let digits = hexString.substringFromIndex(hexString.startIndex.advancedBy(1))
guard Int(digits,radix:16) != nil else{
return nil
}
let red = digits.substringToIndex(digits.startIndex.advancedBy(2))
let green = digits.substringWithRange(Range<String.Index>(start: digits.startIndex.advancedBy(2),
end: digits.startIndex.advancedBy(4)))
let blue = digits.substringWithRange(Range<String.Index>(start:digits.startIndex.advancedBy(4),
end:digits.startIndex.advancedBy(6)))
let redf = CGFloat(Double(Int(red, radix:16)!) / 255.0)
let greenf = CGFloat(Double(Int(green, radix:16)!) / 255.0)
let bluef = CGFloat(Double(Int(blue, radix:16)!) / 255.0)
self.init(red: redf, green: greenf, blue: bluef, alpha: CGFloat(1.0))
}
}
此代码包括字符串格式检查。 如。
let aColor = UIColor(hexString: "#dadada")!
let failed = UIColor(hexString: "123zzzz")
据我所知,我的代码在维护可失败条件的语义和返回可选值方面没有任何缺点。这应该是最好的答案。
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
有一个很好的UIColor类别,其中有许多功能。
用法:
textView.textColor = [UIColor colorWithHexString:textColorHex];
NSLog(@"Text Color Hex: %@", textColorHex);
其中textColorHex有一个形式的@“FFFFFF”没有#符号。
斯威夫特4
你可以像这样在扩展中创建一个非常方便的构造函数:
extension UIColor {
convenience init(hexString: String, alpha: CGFloat = 1.0) {
var hexInt: UInt32 = 0
let scanner = Scanner(string: hexString)
scanner.charactersToBeSkipped = CharacterSet(charactersIn: "#")
scanner.scanHexInt32(&hexInt)
let red = CGFloat((hexInt & 0xff0000) >> 16) / 255.0
let green = CGFloat((hexInt & 0xff00) >> 8) / 255.0
let blue = CGFloat((hexInt & 0xff) >> 0) / 255.0
let alpha = alpha
self.init(red: red, green: green, blue: blue, alpha: alpha)
}
}
以后再用
let color = UIColor(hexString: "#AABBCCDD")
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