我如何从十六进制字符串格式创建一个UIColor,如#00FF00?
当前回答
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];
}
其他回答
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")
斯威夫特2.0:
将此方法添加到VC或扩展UIColor。
func colorWithHexString (hex:String) -> UIColor {
var cString:String = hex.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString
if (cString.hasPrefix("#")) {
cString = (cString as NSString).substringFromIndex(1)
}
if (cString.characters.count != 6) {
return UIColor.grayColor()
}
let rString = (cString as NSString).substringToIndex(2)
let gString = ((cString as NSString).substringFromIndex(2) as NSString).substringToIndex(2)
let bString = ((cString as NSString).substringFromIndex(4) as NSString).substringToIndex(2)
var r:CUnsignedInt = 0, g:CUnsignedInt = 0, b:CUnsignedInt = 0;
NSScanner(string: rString).scanHexInt(&r)
NSScanner(string: gString).scanHexInt(&g)
NSScanner(string: bString).scanHexInt(&b)
return UIColor(red: CGFloat(r) / 255.0, green: CGFloat(g) / 255.0, blue: CGFloat(b) / 255.0, alpha: CGFloat(1))
}
用法:
loginButton.tintColor = self.colorWithHexString("#be1337")
OR
let hexColor = self.colorWithHexString("#be1337")
大多数发布的解决方案使用了Scanner,但至少在现代Swift中你并不真正需要它。相反,你可以简单地使用UInt init和基数16,然后使用基本的二进制操作来获得UIColor组件:
func stringToColor(color: String) -> UIColor {
guard let i = UInt(color, radix: 16) else {
return UIColor.white
}
return UIColor(
red: CGFloat((i & 0xFF0000) >> 16) / 255.0,
green: CGFloat((i & 0xFF00) >> 8) / 255.0,
blue: CGFloat(i & 0xFF) / 255.0,
alpha: 1.0
)
}
这个解决方案期望输入像“FF00FF”,你可能需要删除前面的哈希符号(#),如果你的字符串中有一个。
Swift 5, iOS 14
convenience init(hex: String, alpha: CGFloat = 1.0) {
var hexFormatted: String = hex.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines).uppercased()
if hexFormatted.hasPrefix("#") {
hexFormatted = String(hexFormatted.dropFirst())
}
assert(hexFormatted.count == 6, "Invalid hex code used.")
var rgbValue: UInt64 = 0
Scanner(string: hexFormatted).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)
}
除了颜色,我还喜欢保证alpha,所以我自己写类别
+ (UIColor *) colorWithHex:(int)color {
float red = (color & 0xff000000) >> 24;
float green = (color & 0x00ff0000) >> 16;
float blue = (color & 0x0000ff00) >> 8;
float alpha = (color & 0x000000ff);
return [UIColor colorWithRed:red/255.0 green:green/255.0 blue:blue/255.0 alpha:alpha/255.0];
}
像这样很容易使用
[UIColor colorWithHex:0xFF0000FF]; //Red
[UIColor colorWithHex:0x00FF00FF]; //Green
[UIColor colorWithHex:0x00FF00FF]; //Blue
[UIColor colorWithHex:0x0000007F]; //transparent black
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