我需要绘制4个点并用线性渐变填充该区域,每个点都有不同的颜色。是否可以在HTML5,SVG或任何其他"浏览器" -way中执行此操作?
感谢。
答案 0 :(得分:4)
我在this小提琴
中试验了以下代码<svg height="500" width="500">
<linearGradient id="R">
<stop offset="0" stop-color="red" stop-opacity="1"/>
<stop offset="1" stop-color="white" stop-opacity="0"/>
</linearGradient>
<linearGradient id="G" gradientTransform="rotate(180 0.5 0.5)">
<stop offset="0" stop-color="green" stop-opacity="1"/>
<stop offset="1" stop-color="white" stop-opacity="0"/>
</linearGradient>
<linearGradient id="B" gradientTransform="rotate(270 0.5 0.5)">
<stop offset="0" stop-color="blue" stop-opacity="1"/>
<stop offset="1" stop-color="white" stop-opacity="0"/>
</linearGradient>
<path d="M 100,100 L 300,100 L 200,300 Z" fill="url(#R)"/>
<path d="M 300,100 L 100,100 L 200,300 Z" fill="url(#G)"/>
<path d="M 200,300 L 300,100 L 100,100 Z" fill="url(#B)"/>
</svg>
获得此结果
HTH
编辑我尝试改进并扩展到4分:请参阅updated小提琴。您的问题对我学习SVG结构的基础非常有用。
<svg height="500" width="500">
<linearGradient id="R" gradientTransform="rotate(45 .5 .5)">
<stop offset="0" stop-color="red" stop-opacity="1"/>
<stop offset=".5" stop-color="white" stop-opacity="0"/>
</linearGradient>
<linearGradient id="G" gradientTransform="rotate(135 .5 .5)">
<stop offset="0" stop-color="green" stop-opacity="1"/>
<stop offset=".5" stop-color="white" stop-opacity="0"/>
</linearGradient>
<linearGradient id="B" gradientTransform="rotate(225 .5 .5)">
<stop offset="0" stop-color="blue" stop-opacity="1"/>
<stop offset=".5" stop-color="white" stop-opacity="0"/>
</linearGradient>
<linearGradient id="Y" gradientTransform="rotate(315 .5 .5)">
<stop offset="0" stop-color="yellow" stop-opacity="1"/>
<stop offset=".5" stop-color="white" stop-opacity="0"/>
</linearGradient>
<defs>
<path id="P" d="M 100,100 L 300,100 L 300,300 L 100,300 Z"/>
</defs>
<use xlink:href="#P" fill="url(#R)"/>
<use xlink:href="#P" fill="url(#G)"/>
<use xlink:href="#P" fill="url(#B)"/>
<use xlink:href="#P" fill="url(#Y)"/>
</svg>
值得注意的是,与stop offset="0"
一起玩我们会得到不同的结果......
这里基本: