这是我的d3力量布局: (请运行代码段)
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>
&#13;
这怎么可能?如何在节点之间的链接上应用渐变? 如果问题不明确,请询问。
谢谢!
答案 0 :(得分:5)
结果如下:https://jsfiddle.net/tekh27my/11/
定义部分与@ Cyril的
几乎相同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);
但是在每个刻度线上都需要对x1,y1和x2,y2进行动态更新
所以这是“tick”功能的代码:
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是路径的中间(你可以猜到),因为根据我的计算,这些是单位向量。
gradientVector是一个缩放到0.5的单位向量。
这是单位矢量计算代码:
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);
};
注意:“tick”内部路径的sourceDelta / targetDelta计算与此问题无关。
答案 1 :(得分:2)
您可以执行以下操作并创建渐变并将渐变作为id:
传递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 + ")";
});
工作代码here
希望这有帮助!