Question

我刚刚阅读了这篇帖子Links and Arrowheads to terminate at borders of nodes in D3-但是，我很难用我的示例（我认为实际上）将更简单/不同的弯曲边缘用于我的链接来绘制他的答案（在他的弯曲链接上）。

我一直在研究问题的可重现示例，在过去20至30分钟内显示了力图，但是由于某种原因，该图未出现（即使代码段未引发错误）。尽管只需要固定一小部分，但这不可避免地是一堆代码（重新创建d3力图）。首先，这是代码段：

const svg = d3.select('#mySVG')
const nodesG = svg.select("g.nodes")

var graphs = {
  "nodes": [
    { "name": "Peter", "label": "Person", "id": 1 },
    { "name": "Michael", "label": "Person", "id": 2 },
    { "name": "Neo4j", "label": "Database", "id": 3 },
    { "name": "Graph Database", "label": "Database", "id": 4 }
  ],
  "links": [
    { "source": 1, "target": 2, "type": "KNOWS", "since": 2010 },
    { "source": 1, "target": 3, "type": "FOUNDED" },
    { "source": 2, "target": 3, "type": "WORKS_ON" },
    { "source": 3, "target": 4, "type": "IS_A" }
  ]
}

svg.append('defs').append('marker')
    .attr('id','arrowhead')
    .attr('viewBox','-0 -5 10 10')
    .attr('refX',13)
    .attr('refY',0)
    .attr('orient','auto')
    .attr('markerWidth',13)
    .attr('markerHeight',13)
    .attr('xoverflow','visible')
    .append('svg:path')
    .attr('d', 'M 0,-5 L 10 ,0 L 0,5')
    .attr('fill', '#999')
    .style('stroke','none');

const simulation = d3.forceSimulation()
  .force("link", d3.forceLink().id(d => d.id))
  .force("charge", d3.forceManyBody())
  .force("center", d3.forceCenter(100, 100));

let linksData = graphs.links.map(link => {
  var obj = link;
  obj.source = link.source;
  obj.target = link.target;
  return obj;
})

const links = svg.select("g.links")
  .selectAll("path")
  .data(linksData)
  .enter()
  .append("path")
  .attr('stroke', '#666666')
  .attr('fill', 'transparent')
  .attr("stroke-width", 2)
  .attr('marker-end', 'url(#arrowhead)')

const nodes = nodesG
  .selectAll("g")
  .data(graphs.nodes)
  .enter().append("g")
  .attr("cursor", "pointer")
  .call(d3.drag()
    .on("start", dragstarted)
    .on("drag", dragged)
    .on("end", dragended));

const circles = nodes.append("circle")
  .attr("r", 12)
  .attr("fill", "000000")

nodes.append("title") 
  .text(function(d) { return d.id; });

simulation
  .nodes(graphs.nodes)
  .on("tick", ticked);

simulation.force("link", d3.forceLink().links(linksData)
  .id((d,i) => d.id)
  .distance(150));
  
function ticked() {
  links.attr("d", function(d) {
    var dx = (d.target.x - d.source.x),
        dy = (d.target.y - d.source.y),
        dr = Math.sqrt(dx * dx + dy * dy);
    return "M" + d.source.x + "," + d.source.y + "A" + dr + "," + dr + " 0 0,1 " + d.target.x + "," + d.target.y;
  });

    nodes
        .attr("transform", d => `translate(${d.x}, ${d.y})`);
}

function dragstarted(d) {
  if (!d3.event.active) simulation.alphaTarget(0.3).restart();
  d.fx = d.x;
  d.fy = d.y;
}

function dragged(d) {
  d.fx = d3.event.x;
  d.fy = d3.event.y;
}

function dragended(d) {
  if (!d3.event.active) simulation.alphaTarget(0);
  d.fx = null;
  d.fy = null;
}

<html lang="en">
<head>
    <meta charset="utf-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <meta name="viewport" content="width=device-width, initial-scale=1">
    
    <script src="//d3js.org/d3.v4.min.js" type="text/javascript"></script>

</head>
<body>
<svg id="mySVG" width="500" height="500">
  <g class="links" />
	<g class="nodes" />
</svg>

对于初学者来说，首先解决代码片段以显示图形的任何帮助将不胜感激。

但是，主要问题是箭头进入节点，而我希望没有重叠。这个jsfiddle在箭头和节点之间有一个间隙，我认为看起来更好得多-http://jsfiddle.net/yeQS2/89/-尽管我认为在我的示例中，我希望箭头和节点之间的间隙更大。

我相信这是我需要更新的ticked()函数：

function ticked() {
  links.attr("d", function(d) {
    var dx = (d.target.x - d.source.x),
        dy = (d.target.y - d.source.y),
        dr = Math.sqrt(dx * dx + dy * dy);
    return "M" + d.source.x + "," + d.source.y + "A" + dr + "," + dr + " 0 0,1 " + d.target.x + "," + d.target.y;
  });

    nodes
        .attr("transform", d => `translate(${d.x}, ${d.y})`);
}

我正在努力修复可复制的示例。我为此付出了很多努力，并认为这是很多人在尝试制作美观的d3力布局时会遇到的问题。感谢您提供任何帮助！

编辑：感谢任何帮助使图形生效的人-我进去并使链接上的填充透明，因此仅显示了笔划！

Edit2：不确定是否允许这样做，但这是我的一个大型项目，我将在2天之内悬赏这篇文章，或者，如果可以的话，将尽快悬赏获奖者。

Answer 1

在我的答案here中应用相同的想法。

产生：

<html lang="en">

<head>
  <meta charset="utf-8">
  <meta http-equiv="X-UA-Compatible" content="IE=edge">
  <meta name="viewport" content="width=device-width, initial-scale=1">

  <script src="//d3js.org/d3.v4.min.js" type="text/javascript"></script>

</head>

<body>
  <svg id="mySVG" width="500" height="500">
  <g class="links" />
	<g class="nodes" />
</svg>

  // recalculate and back off the distance
  links.attr("d", function(d) {

    // length of current path
    var pl = this.getTotalLength(),
      // radius of circle plus backoff
      r = (12) + 30, //<-- 12 is your radius 30 is the "back-off" distance
      // position close to where path intercepts circle
      m = this.getPointAtLength(pl - r);

    var dx = m.x - d.source.x,
      dy = m.y - d.source.y,
      dr = Math.sqrt(dx * dx + dy * dy);

    return "M" + d.source.x + "," + d.source.y + "A" + dr + "," + dr + " 0 0,1 " + m.x + "," + m.y;
  });

关键是这个

具有箭头和弯曲边缘的D3力图-缩短链接，以使箭头不会与节点重叠

1 个答案: