单击节点时如何打开子节点的子节点

Question

我有一个四节点tree.I想点击第二个节点想直接打开第四个节点。意思是我想打开子节点的子节点。我正在尝试，但无法获得。

Answer 1

//Toggle children on click.
function click(d) {
   // show is true if children currently hidden
   var show = !d.children; 
   showKids (show, d); // set the visibility of this node's kids

    // if on second level node and we want to show children of children
    // (don't need to same for hide, hiding intermediate node will do the job)
   if (d.depth === 1 && show) {
        var arr = d.children || d._children;
      // loop through kids and set them to visible
      arr.forEach (function(child) {
        showKids (true, child);
      })
   }
   update(d);
}

// showKids replaces simple toggle with a show parameter
function showKids (show, d) {
  if (!show) {
         d._children = d.children;
          d.children = null;
      } else {
          d.children = d._children;
          // blank this line as calling showKids with show=true
          // twice in a row will end up nulling .children and ._children
          // so keep d._children populated.
          //
          // Same doesn't apply to the 'if' condition above as we don't
          // pursue hiding down the tree.
          //d._children = null; 
     }
}

这让您可以执行以下操作：

点击二级节点，显示三级和四级节点
再次点击2级节点，隐藏3级和4级节点
再次点击二级节点，显示三级和四级节点
点击第3级节点，第4级节点隐藏
点击2级节点，隐藏3级节点
再次点击二级节点，显示三级和四级节点

上面的代码可能看起来比实际需要的要复杂一些，但是通过简单的切换步骤 6 会 return 到步骤 4 的状态 - 即仅显示第 3 级。代码中的逻辑确保在展开第 2 级节点时始终显示第 4 级节点，而不管它们之前的状态如何。

单击节点时如何打开子节点的子节点

when click on a node how to open child node of a child

d3.js

c3.js