如果你只需要一个图标，就不需要整个字体集，而且我觉得它作为一个单独的元素更“干净”。因此，出于这个目的，在HTML5中可以直接在文档流中放置SVG。然后你可以在你的.CSS样式表中定义一个类，并使用fill属性访问它的背景色:

工作小提琴: http://jsfiddle.net/qmsj0ez1/

请注意，在示例中，我使用:hover来说明行为;如果您只是想更改“正常”状态的颜色，则应该删除伪类。

是的:)

Surfin' Safari - Blog Archive » CSS Masks WebKit now supports alpha masks in CSS. Masks allow you to overlay the content of a box with a pattern that can be used to knock out portions of that box in the final display. In other words, you can clip to complex shapes based off the alpha of an image. [...] We have introduced new properties to provide Web designers with a lot of control over these masks and how they are applied. The new properties are analogous to the background and border-image properties that already exist. -webkit-mask (background) -webkit-mask-attachment (background-attachment) -webkit-mask-clip (background-clip) -webkit-mask-origin (background-origin) -webkit-mask-image (background-image) -webkit-mask-repeat (background-repeat) -webkit-mask-composite (background-composite) -webkit-mask-box-image (border-image)

/* change image color to white */
filter: invert(100%) sepia(16%) saturate(7463%) hue-rotate(222deg) brightness(119%) contrast(115%);

/* change image color to red */`
filter: invert(16%) sepia(99%) saturate(7404%) hue-rotate(4deg) brightness(95%) contrast(118%);

/* change image color to green */
filter: invert(26%) sepia(89%) saturate(1583%) hue-rotate(95deg) brightness(96%) contrast(106%);

/* change image color to blue */
filter: invert(10%) sepia(90%) saturate(5268%) hue-rotate(245deg) brightness(109%) contrast(155%);

我找到了这个很棒的codedeen示例，您插入十六进制颜色值，它返回所需的过滤器，将此颜色应用到png

CSS过滤器生成器转换从黑色到目标十六进制颜色

例如，我需要PNG的颜色为#1a9790

然后你必须对你的PNG应用下面的过滤器

filter: invert(48%) sepia(13%) saturate(3207%) hue-rotate(130deg) brightness(95%) contrast(80%);

PS:代码依赖是基于这里的MultiplyByZer0的精彩回答:如何仅使用CSS过滤器将黑色转换为任何给定的颜色

所有功劳都归他:鼓掌:

I've been able to do this using SVG filter. You can write a filter that multiplies the color of source image with the color you want to change to. In the code snippet below, flood-color is the color we want to change image color to (which is Red in this case.) feComposite tells the filter how we're processing the color. The formula for feComposite with arithmetic is (k1*i1*i2 + k2*i1 + k3*i2 + k4) where i1 and i2 are input colors for in/in2 accordingly. So specifying only k1=1 means it will do just i1*i2, which means multiplying both input colors together.

注意:这只适用于HTML5，因为它使用内联SVG。但我认为，通过将SVG放在一个单独的文件中，您可能能够在较老的浏览器中实现这一点。我还没有尝试过这种方法。

Here's the snippet: <svg xmlns="http://www.w3.org/2000/svg" version="1.1" width="60" height="90" style="float:left"> <defs> <filter id="colorMask1"> <feFlood flood-color="#ff0000" result="flood" /> <feComposite in="SourceGraphic" in2="flood" operator="arithmetic" k1="1" k2="0" k3="0" k4="0" /> </filter> </defs> <image width="100%" height="100%" xlink:href="http://i.stack.imgur.com/OyP0g.jpg" filter="url(#colorMask1)" /> </svg> <svg xmlns="http://www.w3.org/2000/svg" version="1.1" width="60" height="90" style="float:left"> <defs> <filter id="colorMask2"> <feFlood flood-color="#00ff00" result="flood" /> <feComposite in="SourceGraphic" in2="flood" operator="arithmetic" k1="1" k2="0" k3="0" k4="0" /> </filter> </defs> <image width="100%" height="100%" xlink:href="http://i.stack.imgur.com/OyP0g.jpg" filter="url(#colorMask2)" /> </svg> <svg xmlns="http://www.w3.org/2000/svg" version="1.1" width="60" height="90" style="float:left"> <defs> <filter id="colorMask3"> <feFlood flood-color="#0000ff" result="flood" /> <feComposite in="SourceGraphic" in2="flood" operator="arithmetic" k1="1" k2="0" k3="0" k4="0" /> </filter> </defs> <image width="100%" height="100%" xlink:href="http://i.stack.imgur.com/OyP0g.jpg" filter="url(#colorMask3)" /> </svg>

我需要一个特定的颜色，所以滤镜不适合我。

相反，我创建了一个div，利用CSS多个背景图像和线性梯度函数(它自己创建图像)。如果你使用叠加混合模式，你的实际图像将与生成的“渐变”图像混合，包含你想要的颜色(这里，#BADA55)

.colored-image { background-image: linear-gradient(向右，#BADA55， #BADA55)， url("https://i.imgur.com/lYXT8R6.png"); background-blend-mode:覆盖; background-size:包含; 宽度:200 px; 身高:200 px; ｝ < div class = "的" > < / div >