如何在d3.js中使用多个互连节点进行扩展和折叠

Question

我不得不再次提出这个问题，因为我之前没有找到任何有用的答案，而且我花了很多天时间才找到问题的解决方案，但却无法找到解决方案。

这是我想要实现的目标：

我正在尝试使用此链接代码并尝试在父节点单击上添加展开和折叠功能。

我想要实现的目的是隐藏和显示父节点上的子节点。

我正在使用以下代码：

 // http://blog.thomsonreuters.com/index.php/mobile-patent-suits-graphic-of-the-day/
    var links = [
      // {name: "Microsoft", children: "Amazon", type: "licensing"},
      // {name: "Microsoft", children: "HTC", type: "licensing"},
      {source: "Samsung", target: "Apple", type: "suit"},
      {source: "Motorola", target: "Apple", type: "suit"},
      {source: "Nokia", target: "Apple", type: "resolved"},
      {source: "HTC", target: "Apple", type: "suit"},
      {source: "Kodak", target: "Apple", type: "suit"},
      {source: "Microsoft", target: "Barnes & Noble", type: "suit"},
      {source: "Microsoft", target: "Foxconn", type: "suit"},
      {source: "Oracle", target: "Google", type: "suit"},
      {source: "Apple", target: "HTC", type: "suit"},
      {source: "Microsoft", target: "Inventec", type: "suit"},
      {source: "Samsung", target: "Kodak", type: "resolved"},
      {source: "LG", target: "Kodak", type: "resolved"},
      {source: "RIM", target: "Kodak", type: "suit"},
      {source: "Sony", target: "LG", type: "suit"},
      {source: "Kodak", target: "LG", type: "resolved"},
      {source: "Apple", target: "Nokia", type: "resolved"},
      {source: "Qualcomm", target: "Nokia", type: "resolved"},
      {source: "Apple", target: "Motorola", type: "suit"},
      {source: "Microsoft", target: "Motorola", type: "suit"},
      {source: "Motorola", target: "Microsoft", type: "suit"},
      {source: "Huawei", target: "ZTE", type: "suit"},
      {source: "Ericsson", target: "ZTE", type: "suit"},
      {source: "Kodak", target: "Samsung", type: "resolved"},
      {source: "Apple", target: "Samsung", type: "suit"},
      {source: "Kodak", target: "RIM", type: "suit"},
      {source: "Nokia", target: "Qualcomm", type: "suit"}
    ];
    
    var nodes = {};
    
    // Compute the distinct nodes from the links.
    links.forEach(function(link) {
      link.source = nodes[link.source] || (nodes[link.source] = {name: link.source});
      link.target = nodes[link.target] || (nodes[link.target] = {name: link.target});
    });
    
    var width = 960,
        height = 500;
    
    var force = d3.layout.force()
        .nodes(d3.values(nodes))
        .links(links)
        .size([width, height])
        .linkDistance(60)
        .charge(-300)
        .on("tick", tick)
        .start();
    
    var svg = d3.select("body").append("svg")
        .attr("width", width)
        .attr("height", height);
    
    // Per-type markers, as they don't inherit styles.
    svg.append("defs").selectAll("marker")
        .data(["suit", "licensing", "resolved"])
      .enter().append("marker")
        .attr("id", function(d) { return d; })
        .attr("viewBox", "0 -5 10 10")
        .attr("refX", 15)
        .attr("refY", -1.5)
        .attr("markerWidth", 6)
        .attr("markerHeight", 6)
        .attr("orient", "auto")
      .append("path")
        .attr("d", "M0,-5L10,0L0,5");
    
    var path = svg.append("g").selectAll("path")
        .data(force.links())
      .enter().append("path")
        .attr("class", function(d) { return "link " + d.type; })
        .attr("marker-end", function(d) { return "url(#" + d.type + ")"; });
    
    var circle = svg.append("g").selectAll("circle")
        .data(force.nodes())
      .enter().append("circle")
        .attr("r", 6)
        .call(force.drag)
        .on('click', click);
    
    var text = svg.append("g").selectAll("text")
        .data(force.nodes())
      .enter().append("text")
        .attr("x", 8)
        .attr("y", ".31em")
        .text(function(d) { return d.name; });
    
        function click(d) {
            console.log(d);
          
            
        }
    
    // Use elliptical arc path segments to doubly-encode directionality.
    function tick() {
        
      path.attr("d", linkArc);
      circle.attr("transform", transform);
      text.attr("transform", transform);
    }
    
    function linkArc(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;
    }
    
    function transform(d) {
      return "translate(" + d.x + "," + d.y + ")";
    }
    

.link {
      fill: none;
      stroke: #666;
      stroke-width: 1.5px;
    }
    
    #licensing {
      fill: green;
    }
    
    .link.licensing {
      stroke: green;
    }
    
    .link.resolved {
      stroke-dasharray: 0,2 1;
    }
    
    circle {
      fill: #ccc;
      stroke: #333;
      stroke-width: 1.5px;
    }
    
    text {
      font: 10px sans-serif;
      pointer-events: none;
      text-shadow: 0 1px 0 #fff, 1px 0 0 #fff, 0 -1px 0 #fff, -1px 0 0 #fff;
    }

<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.3.13/d3.min.js"></script>

但是在click（）函数中我没有得到它的子元素名称。那么，我如何获取子元素名称并隐藏并在父节点上显示它们。

由于