Question

请，我正在寻找一种解决方案，不显示根节点并连接到它的线路。我已经尝试附加多个追加（＆＃34; svg：svg＆＃34;）并用 root.children [...] 节点填充那些节点，但我不喜欢这样，因为它对我来说效率低下。

<!DOCTYPE html>
<html>
  <head>
    <meta http-equiv="Content-Type" content="text/html;charset=utf-8"/>
    <link type="text/css" rel="stylesheet" href="style.css"/>
    <script type="text/javascript" src="d3/d3.js"></script>
    <script type="text/javascript" src="d3/d3.layout.js"></script>
    <style type="text/css">


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

.node text {
  font-size: 11px;
}

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

    </style>
  </head>
  <body>
    <div id="body">
      <div id="footer">
        d3.layout.tree
        <div class="hint">click or option-click to expand or collapse</div>
      </div>
    </div>
    <script type="text/javascript">

var m = [20, 120, 20, 120],
    w = 1280 - m[1] - m[3],
    h = 800 - m[0] - m[2],
    i = 0,
    root;

var tree = d3.layout.tree()
    .size([h, w]);

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

var vis = d3.select("#body").append("svg:svg")
    .attr("width", w + m[1] + m[3])
    .attr("height", h + m[0] + m[2])
  .append("svg:g")
    .attr("transform", "translate(" + m[3] + "," + m[0] + ")");

d3.json("flare.json", function(json) {
  root = json;
  root.x0 = h / 2;
  root.y0 = 0;

  function toggleAll(d) {
    if (d.children) {
      d.children.forEach(toggleAll);
      toggle(d);
    }
  }

  // Initialize the display to show a few nodes.
  root.children.forEach(toggleAll);
  toggle(root.children[1]);
  toggle(root.children[1].children[2]);
  toggle(root.children[9]);
  toggle(root.children[9].children[0]);

  update(root);
});

function update(source) {
  var duration = d3.event && d3.event.altKey ? 5000 : 500;

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

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

  // Update the nodes…
  var node = vis.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("svg:g")
      .attr("class", "node")
      .attr("transform", function(d) { return "translate(" + source.y0 + "," + source.x0 + ")"; })
      .on("click", function(d) { toggle(d); update(d); });

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

  nodeEnter.append("svg: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);

  // 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"; });

  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 = vis.selectAll("path.link")
      .data(tree.links(nodes), function(d) { return d.target.id; });

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

  // 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.
function toggle(d) {
  if (d.children) {
    d._children = d.children;
    d.children = null;
  } else {
    d.children = d._children;
    d._children = null;
  }
}

    </script>
  </body>
</html>

Answer 1

您可以弹出您在update(source)中创建的最后一个节点，即根节点。像这样：

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

d3.js树节点布局删除根节点

1 个答案: