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)

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"];

Swift 2.0 - Xcode 7.2

为UIColor添加扩展。

文件-新建- Swift文件-命名。添加以下内容。

extension UIColor {
    convenience init(hexString:String) {
        let hexString:NSString = hexString.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
        let scanner            = NSScanner(string: hexString as String)
        if (hexString.hasPrefix("#")) {
            scanner.scanLocation = 1
        }

        var color:UInt32 = 0
        scanner.scanHexInt(&color)

        let mask = 0x000000FF
        let r = Int(color >> 16) & mask
        let g = Int(color >> 8) & mask
        let b = Int(color) & mask

        let red   = CGFloat(r) / 255.0
        let green = CGFloat(g) / 255.0
        let blue  = CGFloat(b) / 255.0
        self.init(red:red, green:green, blue:blue, alpha:1)
    }

    func toHexString() -> String {
        var r:CGFloat = 0
        var g:CGFloat = 0
        var b:CGFloat = 0
        var a:CGFloat = 0
        getRed(&r, green: &g, blue: &b, alpha: &a)
        let rgb:Int = (Int)(r*255)<<16 | (Int)(g*255)<<8 | (Int)(b*255)<<0
        return NSString(format:"#%06x", rgb) as String
    }        
}

用法:

Ex. Setting Button's color from hexCode.
    override func viewWillAppear(animated: Bool) {
        loginButton.tintColor = UIColor(hexString: " hex code here ")
}

Ex. Converting Button's current color to hex Code.

    override func viewWillAppear(animated: Bool) {
        let hexString = loginButton.tintColor.toHexString()
        print("HEX STRING: \(hexString)")

    }

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

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
    )
  }
}