Question

D3: problems between nodes and links

I created the following jsfiddle to get an idea of what i'm trying to achieve:

enter image description here

I want nodes to over(?) links...

Please help me. I'm sorry for my English :)

Script looks like this:

function myGraph(el) {
this.addNode = function (id) {
    nodes.push({ "id": id });
    update();
}
this.removeNode = function (id) {
    var i = 0;
    var n = findNode(id);
    while (i < links.length) {
        if ((links[i]['source'] == n) || (links[i]['target'] == n)) links.splice(i, 1);
        else i++;
    }
    nodes.splice(findNodeIndex(id), 1);
    update();
}
this.addLink = function (source, target) {
    links.push({ "source": findNode(source), "target": findNode(target) });
    update();
}
var findNode = function (id) {
    for (var i in nodes) { if (nodes[i]["id"] === id) return nodes[i] };
}
var findNodeIndex = function (id) {
    for (var i in nodes) { if (nodes[i]["id"] === id) return i };
}
// set up the D3 svg in the specified element
var w = window.jQuery(el).innerWidth(),
    h = window.jQuery(el).innerHeight();
var svg = this.svg = d3.select(el).append("svg:svg")
    .attr("width", w)
    .attr("height", h);
var force = d3.layout.force()
    .gravity(.05)
    .distance(100)
    .charge(-120)
    .size([w, h]);
var nodes = force.nodes(),
    links = force.links();
var update = function () {
    var link = svg.selectAll(".link")
        .data(links, function (d) { return d.source.id + "-" + d.target.id; })
    link.enter().append("line")
        .attr("class", "link")
    link.exit().remove();
    var node = svg.selectAll(".node")
        .data(nodes, function (d) { return d.id; })
    node.enter().append("g")
        .attr("class", "node")
        .call(force.drag);
    node.append("circle")
        .attr("r", function (d) { return 30; })
        .style("fill", "#EFEFEF")
    node.append("text")
        .attr("text-anchor", "right")
        .text(function (d) { return d.id; });
    force.on("tick", function () {
        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 + ")"; });
    });
    // Restart the force layout.
    force.start();
}
// Make it all go
update();
}
graph = new myGraph("#graph");
// You can do this from the console as much as you like...
graph.addNode("Cause");
graph.addNode("Effect");
graph.addLink("Cause", "Effect");
graph.addNode("A");
graph.addNode("B");
graph.addLink("A", "B");
Was it helpful?

Solution

There's no z-index property in SVG; the elements are rendered in the order in which they are specified. This means for you that all the link elements need to be before the node elements in the generated DOM.

The easiest way to achieve this is to have separate g elements for links and nodes, with the former first.

svg.append("g").attr("class", "links");
svg.append("g").attr("class", "nodes");

Then you can append links like this

var link = svg.select(".links").selectAll(".link")
         .data(links, function (d) { return d.source.id + "-" + d.target.id; })
// etc

and nodes likewise. Complete demo here.

Licensed under: CC-BY-SA with attribution
Not affiliated with StackOverflow
scroll top