Question

我使用此D3图表做了一个节点系列。您可以在下面或jsfiddle中查看我的代码。现在，我想在同一张图表中显示多个节点，如何实现呢？在我的示例中，我做了node1数据，还想添加另一个节点，例如node2系列。

HTML

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width">
  <title>JS Bin</title>
  <script src="https://code.jquery.com/jquery-3.1.0.js"></script>
<script src="//d3js.org/d3.v3.min.js"></script>
</head>
<body>

</body>
</html>

CSS

.node {
    cursor: pointer;
  }

  .overlay{
      background-color:#EEE;
  }

  .node circle {
    fill: #fff;
    stroke: steelblue;
    stroke-width: 1.5px;
  }

  .node text {
    font-size:10px; 
    font-family:sans-serif;
  }

  .link {
    fill: none;
    stroke: #ccc;
    stroke-width: 1.5px;
  }

  .templink {
    fill: none;
    stroke: red;
    stroke-width: 3px;
  }

  .ghostCircle.show{
      display:block;
  }

  .ghostCircle, .activeDrag .ghostCircle{
       display: none;
  }

JS

var flare_data = {
    "name": "Node1",
    "children": [
      {
        "name": "Bob: Child of Top Node1",
        "children": [
          {
            "name": "Son of Bob"
          },
          {
            "name": "Daughter of Bob"
          }
        ]
      },
      {
        "name": "Sally: Child of Top Node1"
      }
    ]
  };


  var margin = {top: 20, right: 120, bottom: 20, left: 120},
    width = 960 - margin.right - margin.left,
    height = 800 - margin.top - margin.bottom;

var i = 0,
    duration = 750,
    root;

var tree = d3.layout.tree()
    .size([height, width]);

var diagonal = d3.svg.diagonal()
    .projection(function(d) { return [d.y, d.x]; });

var svg = d3.select("body").append("svg")
    .attr("width", width + margin.right + margin.left)
    .attr("height", height + margin.top + margin.bottom)
    .append("g")
    .attr("transform", "translate(" + margin.left + "," + margin.top + ")");



  root = flare_data;
  root.x0 = height / 2;
  root.y0 = 0;

  function collapse(d) {
    if (d.children) {
      d._children = d.children;
      d._children.forEach(collapse);
      d.children = null;
    }

  }

  root.children.forEach(collapse);
  update(root);


d3.select(self.frameElement).style("height", "800px");

function update(source) {

  // Compute the new tree layout.
  var nodes = tree.nodes(root).reverse(),
      links = tree.links(nodes);

  // Normalize for fixed-depth.
  nodes.forEach(function(d) { d.y = d.depth * 180; });

  // Update the nodes…
  var node = svg.selectAll("g.node")
      .data(nodes, function(d) {return d.id || (d.id = ++i); });

  // Enter any new nodes at the parent's previous position.
  var nodeEnter = node.enter().append("g")
      .attr("class", "node")
      .attr("transform", function(d) { return "translate(" + source.y0 + "," + source.x0 + ")"; })
      .on("click", click);

  nodeEnter.append("circle")
      .attr("r", 1e-6)
      .style("fill", function(d) { return d._children ? "lightsteelblue" : "#fff"; })
      .append("svg:title")
      .text("Siddhu");

  nodeEnter.append("text")
      .attr("x", function(d) { return d.children || d._children ? -10 : 10; })
      .attr("dy", ".35em")
      .attr("text-anchor", function(d) { return d.children || d._children ? "end" : "start"; })
      .text(function(d) { return d.name; })
      .style("fill-opacity", 1e-6)
      .append("svg:title")
          .html("<b>Siddhu</b>");


  // Transition nodes to their new position.
  var nodeUpdate = node.transition()
      .duration(duration)
      .attr("transform", function(d) { return "translate(" + d.y + "," + d.x + ")"; });

  nodeUpdate.select("circle")
      .attr("r", 4.5)
      .style("fill", function(d) { return d._children ? "lightsteelblue" : "#fff"; })
      .style("fill", function(d) { return (d.name == 'analytics') ? "#f20000" : "#04f100"; });




  nodeUpdate.select("text")
      .style("fill-opacity", 1);

  // Transition exiting nodes to the parent's new position.
  var nodeExit = node.exit().transition()
      .duration(duration)
      .attr("transform", function(d) { return "translate(" + source.y + "," + source.x + ")"; })
      .remove();

  nodeExit.select("circle")
      .attr("r", 1e-6);

  nodeExit.select("text")
      .style("fill-opacity", 1e-6);

  // Update the links…
  var link = svg.selectAll("path.link")
      .data(links, function(d) { return d.target.id; });

  // Enter any new links at the parent's previous position.
  link.enter().insert("path", "g")
      .attr("class", "link")
      .attr("d", function(d) {
        var o = {x: source.x0, y: source.y0};
        return diagonal({source: o, target: o});
      });

  // Transition links to their new position.
  link.transition()
      .duration(duration)
      .attr("d", diagonal);

  // Transition exiting nodes to the parent's new position.
  link.exit().transition()
      .duration(duration)
      .attr("d", function(d) {
        var o = {x: source.x, y: source.y};
        return diagonal({source: o, target: o});
      })
      .remove();

  // Stash the old positions for transition.
  nodes.forEach(function(d) {
    d.x0 = d.x;
    d.y0 = d.y;
  });
}

// Toggle children on click.
function click(d) {
  if (d.children) {
    d._children = d.children;
    d.children = null;
  } else {
    d.children = d._children;
    d._children = null;
  }
  update(d);
}

//Expend all wrote by siddhu
function expand(d){   
    var children = (d.children)?d.children:d._children;
    if (d._children) {        
        d.children = d._children;
        d._children = null;       
    }
    if(children)
      children.forEach(expand);
}

function expandAll(){
    expand(root); 
    update(root);
}
//expandAll();

如何在D3树形图中显示多个父节点？

0 个答案: