Question

我正在关注this 学习d3 js树布局的教程，我正在玩它。在单击子节点时，我正在尝试选择所有祖先节点和连接它们的路径。

我已将教程中的默认click函数更改为看起来像这样的内容。

function click(d) {
  console.log(d.parent);
    while(d.parent)
    { 
      d = d.parent;
      console.log(d.parent);
    }
}

这将逐个选择所有祖先节点，但我如何选择它们之间的所有连接路径？例如：如果我想为所有祖先节点和连接路径着色，我该怎么做？

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) {
  console.log(d.parent);
  while (d.parent) {
    d = d.parent;
    console.log(d.parent);
  }
}

	.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="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.11/d3.min.js"></script>

Answer 1

首先给Ids链接路径，如thsi：

  link.enter().insert("path", "g")
    .attr("class", "link")
    .attr("id", function(d){ return ("link" + d.source.id + "-" + d.target.id)})//unique id

然后为所有节点提供id。

nodeEnter.append("circle")
    .attr("r", 1e-6)
    .attr("id", function(d){return "node" + d.id;})//id of the node.

然后在点击功能中根据ID执行选择：

function click(d) {
  //reset all nodes color
  d3.selectAll("circle").style("fill", "white");//reset all node colors
  d3.selectAll("path").style("stroke", "#c3c3c3");//reset the color for all links
  while (d.parent) {
    d3.selectAll("#node"+d.id).style("fill", "red");//color the node
    if (d.parent != "null")
      d3.selectAll("#link"+d.parent.id + "-" + d.id).style("stroke", "red");//color the path
    d = d.parent;
  }
}

工作代码here

在树布局中选择子节点的所有路径和父节点

1 个答案: