我如何从十六进制字符串格式创建一个UIColor,如#00FF00?
当前回答
有一个很好的帖子是关于如何解决OP的问题,从十六进制字符串中提取UIColor。下面给出的解决方案与其他解决方案不同,因为它支持字符串值,可能包括'0x'或'#'前缀的十六进制字符串表示…(请参阅使用)
这是主要的部分…
- (UIColor *)getUIColorObjectFromHexString:(NSString *)hexStr alpha:(CGFloat)alpha
{
// Convert hex string to an integer
unsigned int hexint = [self intFromHexString:hexStr];
// Create a color object, specifying alpha as well
UIColor *color =
[UIColor colorWithRed:((CGFloat) ((hexint & 0xFF0000) >> 16))/255
green:((CGFloat) ((hexint & 0xFF00) >> 8))/255
blue:((CGFloat) (hexint & 0xFF))/255
alpha:alpha];
return color;
}
助手方法……
- (unsigned int)intFromHexString:(NSString *)hexStr
{
unsigned int hexInt = 0;
// Create scanner
NSScanner *scanner = [NSScanner scannerWithString:hexStr];
// Tell scanner to skip the # character
[scanner setCharactersToBeSkipped:[NSCharacterSet characterSetWithCharactersInString:@"#"]];
// Scan hex value
[scanner scanHexInt:&hexInt];
return hexInt;
}
用法:
NSString *hexStr1 = @"123ABC";
NSString *hexStr2 = @"#123ABC";
NSString *hexStr3 = @"0x123ABC";
UIColor *color1 = [self getUIColorObjectFromHexString:hexStr1 alpha:.9];
NSLog(@"UIColor: %@", color1);
UIColor *color2 = [self getUIColorObjectFromHexString:hexStr2 alpha:.9];
NSLog(@"UIColor: %@", color2);
UIColor *color3 = [self getUIColorObjectFromHexString:hexStr3 alpha:.9];
NSLog(@"UIColor: %@", color3);
完整的参考文章
Swift 2 +
我已经将这个解决方案移植到Swift 2.2。注意,我已经将alpha参数更改为使用默认设置为1.0。我还根据Swift 2.2中的NSScanner类的要求将int类型更新为UInt32。
func colorWithHexString(hexString: String, alpha:CGFloat = 1.0) -> UIColor {
// Convert hex string to an integer
let hexint = Int(self.intFromHexString(hexString))
let red = CGFloat((hexint & 0xff0000) >> 16) / 255.0
let green = CGFloat((hexint & 0xff00) >> 8) / 255.0
let blue = CGFloat((hexint & 0xff) >> 0) / 255.0
// Create color object, specifying alpha as well
let color = UIColor(red: red, green: green, blue: blue, alpha: alpha)
return color
}
func intFromHexString(hexStr: String) -> UInt32 {
var hexInt: UInt32 = 0
// Create scanner
let scanner: NSScanner = NSScanner(string: hexStr)
// Tell scanner to skip the # character
scanner.charactersToBeSkipped = NSCharacterSet(charactersInString: "#")
// Scan hex value
scanner.scanHexInt(&hexInt)
return hexInt
}
斯威夫特 4+
采用与swift 4相同的变化逻辑,
func colorWithHexString(hexString: String, alpha:CGFloat = 1.0) -> UIColor {
// Convert hex string to an integer
let hexint = Int(self.intFromHexString(hexStr: hexString))
let red = CGFloat((hexint & 0xff0000) >> 16) / 255.0
let green = CGFloat((hexint & 0xff00) >> 8) / 255.0
let blue = CGFloat((hexint & 0xff) >> 0) / 255.0
// Create color object, specifying alpha as well
let color = UIColor(red: red, green: green, blue: blue, alpha: alpha)
return color
}
func intFromHexString(hexStr: String) -> UInt32 {
var hexInt: UInt32 = 0
// Create scanner
let scanner: Scanner = Scanner(string: hexStr)
// Tell scanner to skip the # character
scanner.charactersToBeSkipped = CharacterSet(charactersIn: "#")
// Scan hex value
scanner.scanHexInt32(&hexInt)
return hexInt
}
Swift 5 (iOS 13)+
下面显示了在SDK弃用scanHexInt32的情况下工作的更新。我将代码封装到Swift playground文件中。
//: A UIKit based Playground for presenting user interface
import UIKit
import PlaygroundSupport
class MyViewController : UIViewController {
override func loadView() {
let view = UIView()
view.backgroundColor = .white
let label = UILabel()
label.frame = CGRect(x: 150, y: 200, width: 200, height: 20)
label.text = "Hello World!"
label.textColor = colorWithHexString(hexString: "22F728")
view.addSubview(label)
self.view = view
}
func colorWithHexString(hexString: String, alpha:CGFloat = 1.0) -> UIColor {
// Convert hex string to an integer
let hexint = Int(self.intFromHexString(hexStr: hexString))
let red = CGFloat((hexint & 0xff0000) >> 16) / 255.0
let green = CGFloat((hexint & 0xff00) >> 8) / 255.0
let blue = CGFloat((hexint & 0xff) >> 0) / 255.0
// Create color object, specifying alpha as well
let color = UIColor(red: red, green: green, blue: blue, alpha: alpha)
return color
}
func intFromHexString(hexStr: String) -> UInt32 {
var hexInt: UInt32 = 0
// Create scanner
let scanner: Scanner = Scanner(string: hexStr)
// Tell scanner to skip the # character
scanner.charactersToBeSkipped = CharacterSet(charactersIn: "#")
// Scan hex value
hexInt = UInt32(bitPattern: scanner.scanInt32(representation: .hexadecimal) ?? 0)
return hexInt
}
}
// Present the view controller in the Live View window
PlaygroundPage.current.liveView = MyViewController()
彩色十六进制参考 HTML颜色名称和代码 十六进制颜色编码
其他回答
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。
伊森·斯特里德回答的简单例子。一个接受十六进制字符串并返回UIColor的函数。 (你可以输入十六进制字符串格式:#ffffff或ffffff)
例子:
func hexStringToUIColor (hex:String) -> UIColor {
var cString: String = hex.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines).uppercased()
if (cString.hasPrefix("#")) {
if let range = cString.range(of: cString) {
cString = cString.substring(from: cString.index(range.lowerBound, offsetBy: 1))
}
}
if ((cString.characters.count) != 6) {
return UIColor.gray
}
var rgbValue: UInt32 = 0
Scanner(string: cString).scanHexInt32(&rgbValue)
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)
)
}
用法:
var color1 = hexStringToUIColor("#d3d3d3")
你可以使用这个库
https://github.com/burhanuddin353/TFTColor
斯威夫特
UIColor.colorWithRGB(hexString: "FF34AE" alpha: 1.0)
objective - c
[UIColor colorWithRGBHexString:@"FF34AE" alpha:1.0f]
据我所知,没有从十六进制字符串到UIColor(或CGColor)的内置转换。但是,您可以很容易地为此目的编写几个函数—例如,参见iphone开发访问uicolor组件
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。
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