2016-02-20 57 views
0

我想在單擊節點時突出顯示父鏈接。父鏈接意味着被點擊的節點和根之間的所有鏈接。D3樹佈局:如何突出顯示整個路徑

我堅持在點擊節點時選擇鏈接的過程。如何實現這一目標?

的HTML文件:

<!DOCTYPE html> 
<meta charset="utf-8"> 
<style> 
.node circle { 
fill: #fff; 
stroke: steelblue; 
     stroke-width: 3px; 
} 

.node text { 
font: 30px sans-serif; 
} 

.link { 
fill: none; 
stroke: #ccc; 
     stroke-width: 7px; 

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


var margin = {top: 20, right: 20, bottom: 20, left: 20}, 
    width = 1200 - margin.right - margin.left, 
    height = 800 - margin.top - margin.bottom; 

var i = 0, 
    duration = 750, 
    root; 

var tree = d3.layout.tree(); 

var diagonal = d3.svg.diagonal() 
    .projection(function(d) { return [d.x, d.y]; }); 

    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 + ")"); 


    d3.json("parse_tree.json", function(error, parse_tree) { 
      if (error) throw error; 

      root = parse_tree; 
      root.x0 = height/2; 
      root.y0 = 0; 

      tree.size([width, height]); 

      update(root); 
      }); 

function update(source1) { 
    var originalConsole = console; 
    var nodes = tree.nodes(root); 
    var links = tree.links(nodes); 

    nodes.forEach(function(d) { d.y = d.depth * 100; }); 

    // 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(" + source1.y0 + "," + source1.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 ? -20 : 20; }) 
     .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.x + "," + d.y + ")"; }); 

    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(" + source1.y + "," + source1.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: source1.x0, y: source1.y0}; 
       return diagonal({source: o, target: o}); 
       }) 
    .style("stroke-width", "3px") 
     .style("stroke", "green"); 
    // 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: source1.x, y: source1.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; 

      }); 
    for(var k = 0; k < 1000; k++) 
     flag = 0; 
} 

// 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); 
} 


</script> 

JSON文件:

  {"name":"VP", 
      "size":"89", 
     "children": [ 
     {"name":"VBP", 
      "size":"15", 
     "children":[{"name":"are", "size":"38"}] 
     }, 
     {"name":"NP", 
      "size":"83", 
     "children": [ 
     {"name":"DT", 
      "size":"29", 
     "children":[{"name":"a", "size":"53"}] 
     }, 
     {"name":"NN", 
      "size":"50", 
     "children":[{"name":"boy", "size":"99"}] 
     } 
     ] 
     } 
     ] 
     } 

回答

1

這裏的關鍵是links其是物體的陣列,每個具有兩個屬性:源(父節點)和目標(子節點)。有關更多詳細信息,請參見here

在您現有的點擊處理器(function click(d)...),你會通過links要循環檢查,其中links[i].target是您目前點擊節點,每個造型這些鏈接的,你認爲合適的情況下。