2015-11-04 101 views
1

我在d3.js中有一個可摺疊的樹。我的目標是通過「類型」屬性爲節點着色。例如,類型=「str」的節點應填充爲紅色,而類型爲「elem」的節點應爲綠色。我只是無法讓它工作。有人能幫助我嗎。D3可摺疊樹不同節點顏色

這裏是我的代碼:

<!DOCTYPE html> 
<html lang="en"> 
<head> 
<meta charset="utf-8"> 

<title>Tree Collapsible Example</title> 

<style> 

    .node { 
     cursor: pointer; 
    } 

    .node circle { 
     fill: #fff; 
     stroke: steelblue; 
     stroke-width: 2px; 
    } 

    .node text { 
     font: 10px sans-serif; 
    } 

    .link { 
     fill: none; 
     stroke: #ccc; 
     stroke-width: 1.5px; 
    } 

</style> 

</head> 

<body> 

<!-- load the d3.js library --> 
<script src="http://d3js.org/d3.v3.min.js"></script> 

<script> 

// ************** Generate the tree diagram ***************** 
var margin = {top: 20, right: 120, bottom: 20, left: 120}, 
     width = 1000 - 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 + ")"); 

// load the external data 
d3.json("dendrogram02.json", function(error, treeData) { 
    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 * 80; }); 

    // 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 ? 20 : 13; }) 
      .attr("dy", function(d) { return d.children || d._children ? "-.8em" : ".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", 4) 
      .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) { 
    if (d.children) { 
     d._children = d.children; 
     d.children = null; 
    } else { 
     d.children = d._children; 
     d._children = null; 
    } 
    update(d); 
} 

</script> 

</body> 
</html> 

以及相應的以.json:

[ 
    { 
    "name": "1", 
    "type": "sources", 
    "children": [ 
     { 
     "name": "0", 
     "type": "dfasdl", 
     "children": [ 
      { 
      "name": "companies", 
      "type": "seq", 
      "children": [ 
       { 
       "name": "row", 
       "type": "elem", 
       "children": [ 
        { 
        "name": "id", 
        "type": "str" 
        }, 
        { 
        "name": "companyName", 
        "type": "str" 
        }, 
        { 
        "name": "industry", 
        "type": "str" 
        }, 
        { 
        "name": "telephoneCompany", 
        "type": "str" 
        }, 
        { 
        "name": "date_entered", 
        "type": "datetime" 
        } 
       ] 
       } 
      ] 
      }, 
      { 
      "name": "contacts", 
      "type": "seq", 
      "children": [ 
       { 
       "name": "row", 
       "type": "elem", 
       "children": [ 
        { 
        "name": "id", 
        "type": "str" 
        }, 
        { 
        "name": "title", 
        "type": "str" 
        }, 
        { 
        "name": "contactFirstName", 
        "type": "str" 
        }, 
        { 
        "name": "contactLastName", 
        "type": "str" 
        }, 
        { 
        "name": "telephoneUS", 
        "type": "str" 
        } 
       ] 
       } 
      ] 
      }, 
      { 
      "name": "employees", 
      "type": "seq", 
      "children": [ 
       { 
       "name": "row", 
       "type": "elem", 
       "children": [ 
        { 
        "name": "id", 
        "type": "str" 
        }, 
        { 
        "name": "position", 
        "type": "str" 
        }, 
        { 
        "name": "employeeFirstName", 
        "type": "str" 
        }, 
        { 
        "name": "employeeLastName", 
        "type": "str" 
        }, 
        { 
        "name": "telephoneUS", 
        "type": "str" 
        } 
       ] 
       } 
      ] 
      }, 
      { 
      "name": "vendors", 
      "type": "seq", 
      "children": [ 
       { 
       "name": "row", 
       "type": "elem", 
       "children": [ 
        { 
        "name": "id", 
        "type": "str" 
        }, 
        { 
        "name": "type", 
        "type": "str" 
        }, 
        { 
        "name": "vendorFirstName", 
        "type": "str" 
        }, 
        { 
        "name": "vendorLastName", 
        "type": "str" 
        }, 
        { 
        "name": "telephone", 
        "type": "str" 
        } 
       ] 
       } 
      ] 
      } 
     ] 
     } 
    ] 
    } 
] 

回答

3

所有你需要做的是根據您的要求修改

nodeEnter.append("circle") 
     .attr("r", 1e-6) 
     .style("fill", function(d) { return d._children ? "lightsteelblue" : "#fff"; }); 

nodeEnter.append("circle") 
     .attr("r", 1e-6) 
     .style("fill", function(d) { 
      if(d.type == "str") return "red"; 
      if(d.type == "elem") return "green"; 
     }); 

當你設置填充顏色兩次,你也必須爲更新選擇這樣做:

nodeUpdate.select("circle") 
     .attr("r", 4) 
     .style("fill", function(d) { 
      if(d.type == "str") return "red"; 
      if(d.type == "elem") return "green"; 
     }); 
+0

我想這一點,但隨後各界獲得或保持本色... – Johannes

+0

您還需要爲更新選擇做到這一點。我澄清了這一點。 –

+0

啊好的,這個工程!非常感謝! – Johannes