如何在父节点上展开和折叠子节点，请单击d3.js

Question

我正在尝试使用this link code并尝试在父节点点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：

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    // 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 + ")";
    }
    

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.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;
    }

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

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但在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