Question

我正在寻找一种利用this treeview的方法，但是节点开始折叠而不是展开。

这是d3noob片段形式的示例：

var treeData = [
  {
    "name": "Top Level",
    "parent": "null",
    "children": [
      {
        "name": "Level 2: A",
        "parent": "Top Level",
        "children": [
          {
            "name": "Son of A",
            "parent": "Level 2: A"
          },
          {
            "name": "Daughter of A",
            "parent": "Level 2: A"
          }
        ]
      },
      {
        "name": "Level 2: B",
        "parent": "Top Level"
      }
    ]
  }
];


// ************** Generate the tree diagram	 *****************
var margin = {top: 20, right: 120, bottom: 20, left: 120},
	width = 960 - margin.right - margin.left,
	height = 500 - 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 = treeData[0];
root.x0 = height / 2;
root.y0 = 0;
  
update(root);

d3.select(self.frameElement).style("height", "500px");

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"; });

  nodeEnter.append("text")
	  .attr("x", function(d) { return d.children || d._children ? -13 : 13; })
	  .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", 10)
	  .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 = 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);
}

	
	.node {
		cursor: pointer;
	}

	.node circle {
	  fill: #fff;
	  stroke: steelblue;
	  stroke-width: 3px;
	}

	.node text {
	  font: 12px sans-serif;
	}

	.link {
	  fill: none;
	  stroke: #ccc;
	  stroke-width: 2px;
	}

<script src="http://d3js.org/d3.v3.min.js"></script>

任何帮助将不胜感激。

欢呼 KH

Answer 1

您引用的代码段通过将子级放置在树形布局生成器忽略的属性中来隐藏它们：

// Toggle children on click.
function click(d) {
  if (d.children) {
    d._children = d.children;  // copy
    d.children = null;         // remove original
  } else {
    d.children = d._children;  // copy back
    d._children = null;        // remove duplicate
  }
  update(d);  // call the update
}

您可以用相同的方式隐藏除根节点以外的所有内容：

root._children = root.children;
root.children = null;

但是，这可能并不令人满意：所有的孙代在扩展根部时仍会出现：我们没有隐藏它们。因此，我们可以通过一点递归将每个节点的所有子节点移至其_children：

function hideChildren(node) {
    if(node.children) {
        node._children = node.children;
        node.children = null;
        node._children.forEach(hideChildren);
    }
}
hideChildren(root);

这里是updated block。

D3树视图-从折叠的节点开始

1 个答案: