如何在d3.js中返回路径的y坐标?

Jim*_*y C 24 javascript svg d3.js

我有一个图表,我希望代表图形的线在svg区域顶部悬停时在x坐标处显示一个圆圈.该圆应遵循表示曲线的线的路径.问题是我不知道该怎么做.

下面的代码显示了我已经取得了多少成功,它确实在右边x坐标处为文档添加了一个圆圈.现在,我该用什么替换问号?

    svg.on("mousemove", function() {
        d3.select("path.line")
          .style("stroke-width", "2.5px");

         svg.append("svg:circle")
            .attr("cx", Math.floor(event.offsetX-m[1]))
            .attr("cy", ?)
            .attr("r", "10")
            .attr("fill", "red");
    });
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met*_*ion 27

SVG提供了一个本机函数.getPointAtLength(),该函数返回路径的x和y值,无论你在它传递的长度是多少.

您需要遍历线的长度,直到找到相应的y位置.以下是你在D3中的表现:

var svg = d3.select("#line").append("svg")
var path = 
    svg.append("path")
      .attr("d", "M0,168L28,95.99999999999997L56,192L84,71.99999999999997L112,120L140,192L168,240L196,168L224,48L252,24L280,192L308,120L336,24L364,168L392,95.99999999999997L420,168L448,95.99999999999997L476,192L504,71.99999999999997L532,120L560,192L588,216L616,168L644,48L672,24L700,192L728,120L756,24L784,192L812,71.99999999999997")
      .attr("fill", "none")
      .attr("stroke", "black");

var circle = 
    svg.append("circle")
      .attr("cx", 100)
      .attr("cy", 350)
      .attr("r", 3)
      .attr("fill", "red");

var pathEl = path.node();
var pathLength = pathEl.getTotalLength();
var BBox = pathEl.getBBox();
var scale = pathLength/BBox.width;
var offsetLeft = document.getElementById("line").offsetLeft;
var randomizeButton = d3.select("button");

svg.on("mousemove", function() {
  var x = d3.event.pageX - offsetLeft; 
  var beginning = x, end = pathLength, target;
  while (true) {
    target = Math.floor((beginning + end) / 2);
    pos = pathEl.getPointAtLength(target);
    if ((target === end || target === beginning) && pos.x !== x) {
        break;
    }
    if (pos.x > x)      end = target;
    else if (pos.x < x) beginning = target;
    else                break; //position found
  }
  circle
    .attr("opacity", 1)
    .attr("cx", x)
    .attr("cy", pos.y);
});
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你可以在这里看到一个演示:http://bl.ocks.org/3824661

  • @Snowfish想象一下,你在∩∩∩---开头有一条宽曲线的线,当你在屏幕中间时,你将处于屏幕宽度的50%.然而,该线可能是其长度的75%,但在您计算所有候选位置之前您不确定(例如,线的长度将永远不会小于屏幕的宽度). (2认同)