CSS3 Webkit-过滤器属性 [英] CSS3 Webkit-filter Property

问题描述

我有一个图像是纯红色，＃FF0000。我的目标是将它更改为＃1100FF。

I have got an image that has got plain red color, #FF0000. My target is to change it to #1100FF.

只是几个星期前我知道我们在CSS3有着色/过滤功能。它的工作相当不错，但图像的最终颜色不是＃1100FF。我通过Chrome工具Awesome颜色选择器检查颜色

Just couple of weeks ago I get to know that we have tinting/filtering capability in CSS3. It works pretty good however final color of the Image is not #1100FF. I check color through Chrome Tool 'Awesome Color Picker'

那么我该怎么做是你一定要问的。

So how I do it is what you must be asking.

function RGB_TO_HSV($R, $G, $B)    // RGB values:    0-255, 0-255, 0-255
                    {                                // HSV values:    0-360, 0-100, 0-100
                        $HSL = array();
                        // Convert the RGB byte-values to percentages
                        $R = ($R / 255);
                        $G = ($G / 255);
                        $B = ($B / 255);

                        $maxRGB = max($R, $G, $B);
                        $minRGB = min($R, $G, $B);
                        $chroma = $maxRGB - $minRGB;

                        $computedV = 100 * $maxRGB;

                        if ($chroma == 0)
                        {
                            $HSL['H']= "0deg";
                            $HSL['S']= "0%";
                            $HSL['V']= (3*$computedV)."%";
                            return $HSL;
                        }
                        $computedS = 100 * ($chroma / $maxRGB);

                        if ($R == $minRGB)
                            $h = 3 - (($G - $B) / $chroma);
                        elseif ($B == $minRGB)
                            $h = 1 - (($R - $G) / $chroma);
                        else // $G == $minRGB
                            $h = 5 - (($B - $R) / $chroma);

                        $computedH = 60 * $h;
                        $HSL['H'] = $computedH."deg";
                        $HSL['S'] = $computedS."%";
                        $HSL['V'] = $computedV."%";

                        return $HSL;
                    }

然后在CSS中，我这样做：

Then In CSS, I do this:

img {-webkit-filter: hue-rotate(<?=$HSL['H']?>) saturate(<?=$HSL['S']?>) brightness(<?=$HSL['V']?>);}

现在请告诉我如何实现所需的颜色？我需要添加或减去原始颜色（即＃FF0000）以获得所需的颜色或我使用的代码不正确或什么？

Now please tell me how I can achieve the required color? Do I need to add or subtract the original color (ie. #FF0000) to get the desired color or the code I am using is not correct or what?

输出颜色是：＃0F2DFF

Output Color is : #0F2DFF

编辑：图像颜色没有限制＃FF0000。

There is no restriction that color of the image is #FF0000. I can change it as long as I can tint the image to the any color accurately.

推荐答案

我同意Joram的看法。它不能按预期工作。

I agree with Joram. It is not working as expected.

我最好的朋友是CSS形状技术。我在css和颜色的形状，以任何我喜欢的，它是像素完美的。虽然它需要一些时间，但它是值得的，很多效率比浪费时间过滤技术。

My best friend was CSS shapes techniques. I made shapes in the css and color it to whatever I liked and it was pixel perfect. Though it take some time but it was worth it and lot effective than wasting time on filtering techniques.

我厌倦了半熟的标准。 ：（

I am tired of half-baked standards. : (

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