我尝试绘制厚Bezier线(用于自定义Sankey图).我使用SVG路径,以贝塞尔曲线的形式C x1 y1, x2 y2, x y.我使用stroke而不是fill,因此它们具有恒定的宽度(并且可以表示流).
如果线条很薄或垂直差异相对较小,它的效果非常好.然而,如果它们非常厚,我会得到一些讨厌的文物(看起来像角) - 请参见下图中的右下曲线:
资料来源:http://jsfiddle.net/stared/83jr5fub/
有没有办法避免伪影,即:
x1或右侧没有任何东西x,stroke-width?我认为在你的情况下(使用给定的路径),他的最佳解决方案是关闭路径,并使用其填充属性.
要做到这一点,你必须做出lineTo(0, strokeWidth)你的结束BezierCurveTo,然后重新绘制贝兹曲线中的其他方式:
var svg = d3.select("#chart");
var data = [
{t: 5, dy: 10},
{t: 5, dy: 20},
{t: 5, dy: 40},
{t: 20, dy: 10},
{t: 20, dy: 20},
{t: 20, dy: 40},
{t: 50, dy: 10},
{t: 50, dy: 20},
{t: 50, dy: 40},
];
var ctrl = 10;
var dx = 40;
var spacing = 100;
var colors = d3.scale.category10();
svg
.attr("width", 4 * spacing)
.attr("height", 4 * spacing);
svg.selectAll("path")
.data(data)
.enter()
.append("path")
.attr("d", function (d, i) {
var x1 = spacing + spacing * (i % 3);
var y1 = spacing + spacing * Math.floor(i / 3);
return "M" + x1 + "," + y1 +
"c" + ctrl + "," + 0 +
" " + (dx - ctrl) + "," + d.dy +
" " + dx + "," + d.dy +
// move down for the wanted width
"l" + (0) + "," + (d.t) +
// negate all values
"c" + (ctrl * -1) + "," + 0 +
" " + ((dx - ctrl) * -1) + "," + (d.dy * -1) +
" " + (dx * -1) + "," + (d.dy * -1);
})
.style("fill", colors(0))Run Code Online (Sandbox Code Playgroud)
<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.11/d3.min.js"></script>
<svg id="chart"></svg>Run Code Online (Sandbox Code Playgroud)
因为这里价值超过1万字的动画显示了正在发生的事情以及为什么它不能被称为浏览器错误:
@keyframes dash {
from {
stroke-dashoffset: -10%;
}
to {
stroke-dashoffset: 90%;
}
}
@-webkit-keyframes dash {
from {
stroke-dashoffset: -10%;
}
to {
stroke-dashoffset: 90%;
}
}
#dashed{
animation : dash 12s linear infinite;
}Run Code Online (Sandbox Code Playgroud)
<svg height="200" width="200" id="chart" viewBox="290 260 100 100">
<path id="dashed" style="fill: none; stroke: rgb(31, 119, 180); stroke-width: 50; stroke-dasharray: 3, 3;" d="M300,300c10,0 30,40 40,40"></path>
<path style="fill: none; stroke: black;" d="M300,300c10,0 30,40 40,40">
</path></svg>Run Code Online (Sandbox Code Playgroud)