我如何从十六进制字符串格式创建一个UIColor,如#00FF00?


当前回答

使用这个类别:

在文件UIColor+Hexadecimal.h中

#import <UIKit/UIKit.h>

@interface UIColor(Hexadecimal)

+ (UIColor *)colorWithHexString:(NSString *)hexString;

@end

在文件UIColor+Hexadecimal.m中

#import "UIColor+Hexadecimal.h"

@implementation UIColor(Hexadecimal)

+ (UIColor *)colorWithHexString:(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];
}

@end

你想在课堂上使用它:

#import "UIColor+Hexadecimal.h"

and:

[UIColor colorWithHexString:@"#6e4b4b"];

其他回答

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。

下面是Swift 1.2版本,作为UIColor的扩展。这允许你这样做

let redColor = UIColor(hex: "#FF0000")

我觉得这是最自然的做法。

extension UIColor {
  // Initialiser for strings of format '#_RED_GREEN_BLUE_'
  convenience init(hex: String) {
    let redRange    = Range<String.Index>(start: hex.startIndex.advancedBy(1), end: hex.startIndex.advancedBy(3))
    let greenRange  = Range<String.Index>(start: hex.startIndex.advancedBy(3), end: hex.startIndex.advancedBy(5))
    let blueRange   = Range<String.Index>(start: hex.startIndex.advancedBy(5), end: hex.startIndex.advancedBy(7))

    var red     : UInt32 = 0
    var green   : UInt32 = 0
    var blue    : UInt32 = 0

    NSScanner(string: hex.substringWithRange(redRange)).scanHexInt(&red)
    NSScanner(string: hex.substringWithRange(greenRange)).scanHexInt(&green)
    NSScanner(string: hex.substringWithRange(blueRange)).scanHexInt(&blue)

    self.init(
      red: CGFloat(red) / 255,
      green: CGFloat(green) / 255,
      blue: CGFloat(blue) / 255,
      alpha: 1
    )
  }
}

我发现最简单的方法是使用宏。只要把它包括在你的标题中,它就可以在你的整个项目中使用。

#define UIColorFromRGB(rgbValue) [UIColor colorWithRed:((float)((rgbValue & 0xFF0000) >> 16))/255.0 green:((float)((rgbValue & 0xFF00) >> 8))/255.0 blue:((float)(rgbValue & 0xFF))/255.0 alpha:1.0]

Uicolor宏十六进制值

这段代码的格式化版本:

#define UIColorFromRGB(rgbValue) \
[UIColor colorWithRed:((float)((rgbValue & 0xFF0000) >> 16))/255.0 \
                green:((float)((rgbValue & 0x00FF00) >>  8))/255.0 \
                 blue:((float)((rgbValue & 0x0000FF) >>  0))/255.0 \
                alpha:1.0]

用法:

label.textColor = UIColorFromRGB(0xBC1128);

迅速:

static func UIColorFromRGB(_ rgbValue: Int) -> UIColor! {
    return UIColor(
        red: CGFloat((Float((rgbValue & 0xff0000) >> 16)) / 255.0),
        green: CGFloat((Float((rgbValue & 0x00ff00) >> 8)) / 255.0),
        blue: CGFloat((Float((rgbValue & 0x0000ff) >> 0)) / 255.0),
        alpha: 1.0)
}

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")

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]