我正在使用以下 Javascript + SVG + D3 代码。
http://bl.ocks.org/1095795 http://bl.ocks.org/1095795
代码在这里：
https://gist.github.com/1095727 https://gist.github.com/1095727
简而言之，我需要为每个节点提供唯一的图像（而不是现在拉动的重复的笑脸图像），但我不够精明，无法实现这一点。任何帮助，将不胜感激。
<!DOCTYPE html>
<html>
<head>
<script type="text/javascript" src="http://mbostock.github.com/d3/d3.js"></script>
<script type="text/javascript" src="http://mbostock.github.com/d3/d3.layout.js"></script>
<script type="text/javascript" src="http://mbostock.github.com/d3/d3.geom.js"></script>
<style type="text/css">
.link { stroke: #ccc; }
.nodetext { pointer-events: none; font: 10px sans-serif; }
</style>
</head>
<body>
<script type="text/javascript">
var w = 960,
h = 500
var nodes = [],
links = [];
var vis = d3.select("body").append("svg:svg")
.attr("width", w)
.attr("height", h);
var force = self.force = d3.layout.force()
.nodes(nodes)
.links(links)
.gravity(.05)
.distance(100)
.charge(-100)
.size([w, h]);
force.on("tick", function() {
var node = vis.selectAll("g.node")
.data(nodes, function(d) { return d.id;} )
var link = vis.selectAll("line.link")
.data(links, function(d) { return d.source.id + ',' + d.target.id})
link.attr("x1", function(d) { return d.source.x; })
.attr("y1", function(d) { return d.source.y; })
.attr("x2", function(d) { return d.target.x; })
.attr("y2", function(d) { return d.target.y; });
node.attr("transform", function(d) { return "translate(" + d.x + "," + d.y + ")"; });
});
function recalc() {
var link = vis.selectAll("line.link")
.data(links, function(l) { return l.source.id + '-' + l.target.id; });
link.enter().append("svg:line")
.attr("class", "link")
.attr("x1", function(d) { return d.source.x; })
.attr("y1", function(d) { return d.source.y; })
.attr("x2", function(d) { return d.target.x; })
.attr("y2", function(d) { return d.target.y; });
link.exit().remove();
var node = vis.selectAll("g.node")
.data(nodes, function(d) { return d.dpid;}).call(force.drag);
var nodeEnter = node.enter().append("svg:g")
.attr("class", "node")
.call(force.drag);
nodeEnter.append("svg:image")
.attr("class", "circle")
.attr("xlink:href", "https://d3nwyuy0nl342s.cloudfront.net/images/icons/public.png")
.attr("x", "-8px")
.attr("y", "-8px")
.attr("width", "16px")
.attr("height", "16px");
nodeEnter.append("svg:text")
.attr("class", "nodetext")
.attr("dx", 12)
.attr("dy", ".35em")
.text(function(d) { return d.id });
node.exit().remove();
force.start();
}
/* Scenario */
/* step 1: add three nodes and three links */
function step1() {
var nA = {id: 'aaa'};
var nB = {id: 'bbb'};
var nC = {id: 'ccc'};
nodes.push(nA);
nodes.push(nB);
nodes.push(nC);
var lAB = {source: nA, target: nB};
var lAC = {source: nA, target: nC};
var lBC = {source: nB, target: nC};
links.push(lAB );
links.push(lAC);
links.push(lBC);
recalc();
}
/* step 2: node B disappears with links */
function step2() {
nodes = nodes.filter(function(n) { return n.id !== 'bbb'; });
links = links.filter(function(l) { return (l.source.id !== 'bbb' && l.target.id !== 'bbb'); });
recalc();
}
/* step 3: node B reappears with links */
function step3() {
var nB = {id: 'bbb'};
nodes.push(nB);
/* find exiting nodes for links */
var nA = nodes.filter(function(n) { return n.id === 'aaa'; })[0];
var nC = nodes.filter(function(n) { return n.id === 'ccc'; })[0];
var lAB = {source: nA, target: nB};
var lBC = {source: nB, target: nC};
links.push(lAB);
links.push(lBC);
recalc();
}
window.setTimeout(step1, 2000);
window.setTimeout(step2, 4000);
window.setTimeout(step3, 6000);
force.start();
recalc();
</script>
</body>
</html>
