swift - 在 swift 中使用颜色的 RGB 值或十六进制值更改 View 的颜色

Question

我在网上得到了下面的代码，我在我的项目中使用了这段代码来改变背景的颜色。

// Creates a UIColor from a Hex string.

func colorWithHexString (hex:String) -> UIColor {
    var cString:String = hex.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString

    if (cString.hasPrefix("#")) {
        cString = (cString as NSString).substringFromIndex(1)
    }

    if (countElements(cString) != 6) {
        return UIColor.grayColor()
    }

    var rString = (cString as NSString).substringToIndex(2)
    var gString = ((cString as NSString).substringFromIndex(2) as NSString).substringToIndex(2)
    var 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))
}

当我如下调用上述函数来改变背景颜色时:

lbl91.backgroundColor = colorWithHexString(hex: 0x209624);

它给我以下错误:

cannot convert the expression's type '()' to type integerliteralconvertible

Answer 1

函数需要这样的参数:

colorWithHexString("#ff00dd")

或者没有标签

colorWithHexString("ff00dd")