D3js force layout - line with gradient between nodes

Question

This is my d3 force layout: (Please run the code snippet)

var width = 600,
    height = 600;

var svg = d3.select('body').append('svg')
    .attr('width', width)
    .attr('height', height);

var color = d3.scale.category20();

var dataNodes = [
    { x:   width/3, y:   height/3 , group: 0, color: 'blue'},
    { x: 2*width/3, y:   height/3, group: 1, color: 'red' },
    { x:   width/2, y: 2*height/3, group: 2, color: 'green'}
];

var dataLinks = [
  { source: 0, target: 1},
  { source: 1, target: 2},
  { source: 2, target: 0}
];

var force = d3.layout.force()
    .charge(-400)
    .linkDistance(height/2)
    .size([width, height])
    .linkStrength(1.3)
    .friction(0.8)
    .gravity(0.9);

force
    .nodes(dataNodes)
    .links(dataLinks)
    .start();

var link = svg.selectAll(".link")
      .data(dataLinks)
    .enter().append("line")
      .attr("class", "link");

var node = svg.selectAll(".node")
    .data(dataNodes)
  .enter().append("circle")
    .attr("class", function(d){ return "node " + d.color})
    .attr("r", width/20)
    .call(force.drag);

node.append("title")
      .text(function(d) { return d.color; });


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("cx", function(d) { return d.x; })
        .attr("cy", function(d) { return d.y; });
});

.node {
    fill: #ccc;
    stroke: #fff;
    stroke-width: 0;
}
.node.blue {
    fill: blue;
}
.node.red {
    fill: red;
}
.node.green {
    fill: green;
}

.link {
    fill: none;         
    stroke: black;
    stroke-width: 20px;
}

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

This is what i want to achieve:

How is this possible ? How can i apply a gradient on the links between the nodes ? If something is not clear please ask.

Thanks !

I don't know if you've solved the issue but I came across the same problem today and i think got the right code for it. Check my answer below. — eko, Jun 06 '16 at 07:40

eko · Answer 1 · 2016-06-06T07:49:20.717

Here's the result: https://jsfiddle.net/tekh27my/11/

The definition part is pretty much same with @Cyril's

var gradient = d3.select("svg").append("defs")
    .append("linearGradient")
    .attr("id", "gradient")
    .attr("spreadMethod", "pad");
  //start color white
  gradient.append("stop")
    .attr("offset", "0%")
    .attr("stop-color", "red")
    .attr("stop-opacity", 1);
  //end color steel blue
    gradient.append("stop")
    .attr("offset", "100%")
    .attr("stop-color", "green")
    .attr("stop-opacity", 1);

But at every tick there needs to be a dynamic update on the x1,y1 and x2,y2

So this is the code for "tick" function:

var linkVector = new Vector2(d.target.x-d.source.x,d.target.y-d.source.y).getUnitVector();
var perpVector = linkVector.perpendicularClockwise().scale(radius);
var gradientVector = linkVector.scale(0.5);


gradient
    .attr("x1", 0.5-gradientVector.X)
    .attr("y1", 0.5-gradientVector.Y)
    .attr("x2", 0.5+gradientVector.X)
    .attr("y2", 0.5+gradientVector.Y);

0.5 is the middle of the path (as you can guess) since these are unit vectors according to my calculations.

gradientVector is a unit vector scaled to 0.5.

And here's the unit vector calculation code:

var Vector2 = function(x,y) {
  this.magnitude = Math.sqrt(x*x+y*y);
  this.X = x;
  this.Y = y;
};

Vector2.prototype.perpendicularClockwise = function(){
  return new Vector2(-this.Y, this.X);
};

Vector2.prototype.perpendicularCounterClockwise = function(){
  return new Vector2(this.Y, -this.X);
};

Vector2.prototype.getUnitVector = function(){
  return new Vector2(this.X/this.magnitude, this.Y/this.magnitude);
};

Vector2.prototype.scale = function(ratio){
  return new Vector2(ratio*this.X, ratio*this.Y);
};

Note: The sourceDelta/targetDelta calculation for the path inside the "tick" is irrelevant for this question.

score 2 · Answer 2 · answered Feb 07 '16 at 11:01

2

You can do something like this and create gradient and pass the gradient as the id:

var link = svg.selectAll(".link")
      .data(dataLinks)
    .enter().append("line")
      .attr("class", "link")
      .style("stroke",function(d){
          var id = "S"+d.source.index +"T" + d.target.index;
          var gradient1 = defs.append("linearGradient").attr("id",  id);
          gradient1.append("stop").attr("offset", "0%").attr("stop-color", d.target.color);
          gradient1.append("stop").attr("offset", "100%").attr("stop-color", d.source.color);
          return "url(#" + id + ")";
      });

working code here

hope this helps!

answered Feb 07 '16 at 11:01

Cyril Cherian

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when you turn the node around, the gradient does not seem to update correctly – ee2Dev Feb 07 '16 at 15:18
Thanks for your answer. But sometimes the "direction" of the gradient is not relative to the position of the nodes like ee2Dev mentioned it. Is there a way to change the direction of the gradient while turning the nodes ? – Mangocrack Feb 07 '16 at 15:30
Thanks for pointing that out...yes on dragging the nodes the gradient is getting reset...sry not able to fix it.. – Cyril Cherian Feb 08 '16 at 10:05
@Cyril I've added some code to your original answer, just leaving a note if you are still into the gradient stuff – eko Jun 06 '16 at 07:39

D3js force layout - line with gradient between nodes

2 Answers2