给定一个透明的PNG显示一个简单的形状在白色,它是有可能以某种方式改变这通过CSS的颜色?某种叠加还是什么?
当前回答
我在谷歌上找到了这个,我发现最适合我的工作…
HTML
<div class="img"></div>
CSS
.img {
background-color: red;
width: 60px;
height: 60px;
-webkit-mask-image: url('http://i.stack.imgur.com/gZvK4.png');
}
http://jsfiddle.net/a63b0exm/
其他回答
当将一张图片从黑到白,或从白到黑时,色调旋转滤镜不起作用,因为黑色和白色在技术上不是颜色。相反,黑白颜色的变化(从黑到白或反之)必须使用invert filter属性来完成。
.img1 { 过滤器:反转(100%); }
为了字面上改变颜色,你可以使用-webkit-filter来合并CSS转换,当有事情发生时,你可以调用你选择的-webkit-filter。例如:
img {
-webkit-filter:grayscale(0%);
transition: -webkit-filter .3s linear;
}
img:hover
{
-webkit-filter:grayscale(75%);
}
你可以使用-webkit-filter和filter来使用过滤器: 过滤器对于浏览器来说相对较新,但根据以下CanIUse表:https://caniuse.com/#feat=css-filters,超过90%的浏览器支持过滤器
你可以将图像更改为灰度,深褐色和更多(请看示例)。
所以你现在可以用滤镜改变PNG文件的颜色。
body { background-color:#03030a; min-width: 800px; min-height: 400px } img { width:20%; float:left; margin:0; } /*Filter styles*/ .saturate { filter: saturate(3); } .grayscale { filter: grayscale(100%); } .contrast { filter: contrast(160%); } .brightness { filter: brightness(0.25); } .blur { filter: blur(3px); } .invert { filter: invert(100%); } .sepia { filter: sepia(100%); } .huerotate { filter: hue-rotate(180deg); } .rss.opacity { filter: opacity(50%); } <!--- img src http://upload.wikimedia.org/wikipedia/commons/thumb/e/ec/Mona_Lisa%2C_by_Leonardo_da_Vinci%2C_from_C2RMF_retouched.jpg/500px-Mona_Lisa%2C_by_Leonardo_da_Vinci%2C_from_C2RMF_retouched.jpg --> <img alt="Mona Lisa" src="https://images.pexels.com/photos/40997/mona-lisa-leonardo-da-vinci-la-gioconda-oil-painting-40997.jpeg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260" title="original"> <img alt="Mona Lisa" src="https://images.pexels.com/photos/40997/mona-lisa-leonardo-da-vinci-la-gioconda-oil-painting-40997.jpeg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260" title="saturate" class="saturate"> <img alt="Mona Lisa" src="https://images.pexels.com/photos/40997/mona-lisa-leonardo-da-vinci-la-gioconda-oil-painting-40997.jpeg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260" title="grayscale" class="grayscale"> <img alt="Mona Lisa" src="https://images.pexels.com/photos/40997/mona-lisa-leonardo-da-vinci-la-gioconda-oil-painting-40997.jpeg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260" title="contrast" class="contrast"> <img alt="Mona Lisa" src="https://images.pexels.com/photos/40997/mona-lisa-leonardo-da-vinci-la-gioconda-oil-painting-40997.jpeg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260" title="brightness" class="brightness"> <img alt="Mona Lisa" src="https://images.pexels.com/photos/40997/mona-lisa-leonardo-da-vinci-la-gioconda-oil-painting-40997.jpeg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260" title="blur" class="blur"> <img alt="Mona Lisa" src="https://images.pexels.com/photos/40997/mona-lisa-leonardo-da-vinci-la-gioconda-oil-painting-40997.jpeg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260" title="invert" class="invert"> <img alt="Mona Lisa" src="https://images.pexels.com/photos/40997/mona-lisa-leonardo-da-vinci-la-gioconda-oil-painting-40997.jpeg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260" title="sepia" class="sepia"> <img alt="Mona Lisa" src="https://images.pexels.com/photos/40997/mona-lisa-leonardo-da-vinci-la-gioconda-oil-painting-40997.jpeg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260" title="huerotate" class="huerotate"> <img alt="Mona Lisa" src="https://images.pexels.com/photos/40997/mona-lisa-leonardo-da-vinci-la-gioconda-oil-painting-40997.jpeg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260" title="opacity" class="rss opacity">
源
使用这个很棒的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,我需要我的png颜色为#EF8C57
然后你必须对你的PNG应用下面的过滤器 结果:
filter: invert(76%) sepia(30%) saturate(3461%) hue-rotate(321deg) brightness(98%) contrast(91%);
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>
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