目前,我正在使用FCC的一个项目National Contiguity
基本上我能够弄清楚如何渲染标志精灵并连接节点及其链接。
我唯一遇到的问题是如何实现类似于FCC example的行为。具体来说,我正在尝试让节点像示例中那样排斥边界。
然而,我的行为并不是这样(事实上,它非常集中,但我不确切地知道d3.js Force Documents V4中应该查看哪些属性。相反,似乎虽然节点和链接在边界附近停止,但它在那个点停止
const width = w - (margin.left + margin.right);
const height = h - (margin.top + margin.bottom);
let flagNodes = d3.select("#canvas")
.append("div")
.classed("flag-nodes",true)
let svg = d3.select("#canvas")
.append("svg")
.attr("id","chart")
.attr("width", w)
.attr("height", h)
let chart = svg.append("g")
.classed("display", true)
.attr("transform", "translate(" + margin.left + "," + margin.top + ")");
let simulation = d3.forceSimulation()
.force("link", d3.forceLink().id(function(d,i) {
return i;
}))
.force("charge", d3.forceManyBody().strength(-60).distanceMax(50).distanceMin(5))
.force("center", d3.forceCenter(width/2, height/2))
.force("collide", d3.forceCollide().radius(35))
// .force("centering", d3.forceCenter(,height))
// .force("position", d3.forceX(0).strength(.01))
// .force("position", d3.forceY(-18))
let link = chart.append("g")
.classed("links",true)
.selectAll("line")
.data(data.links)
.enter()
.append("line")
simulation
.nodes(data.nodes)
.on("tick", ticked);
simulation
.force("link")
.links(data.links);
let node = flagNodes.selectAll(".flag-nodes")
.data(data.nodes)
.enter()
.append("div")
.attr("class", function(d,i){
return `flag flag-${d.code}`
})
.call(d3.drag()
.on("start", dragstarted)
.on("drag", dragged)
.on("end", dragended)
)
node.append("title")
.text(function(d) { return d.country; });
d3.forceX(width)
//functions provided by D3.js
//
function ticked() {
node
.style("left", function(d) {
let xlimit = Math.max(radius, Math.min(width - radius, d.x))
return (xlimit) + 'px'
})
.style("top", function(d) {
let ylimit = Math.max(radius, Math.min(height - radius, d.y))
return (ylimit - 2) + 'px'
});
link
.attr("x1", function(d) {
let x1 = Math.max(radius, Math.min(width - radius, d.source.x))
return x1;
})
.attr("y1", function(d) {
let y1 = Math.max(radius, Math.min(height - radius, d.source.y))
return y1
})
.attr("x2", function(d) {
let x2 = Math.max(radius, Math.min(width - radius, d.target.x))
return x2;
})
.attr("y2", function(d) {
let y2 = Math.max(radius, Math.min(height - radius, d.target.y))
return y2
});
}
function dragstarted(d) {
if (!d3.event.active) simulation.alphaTarget(0.3).restart();
d.fx = d.x;
d.fy = d.y;
}
function dragged(d) {
d.fx = d3.event.x;
d.fy = d3.event.y;
}
function dragended(d) {
if (!d3.event.active) simulation.alphaTarget(0);
d.fx = null;
d.fy = null;
}
答案 0 :(得分:1)
在每个刻度线上,在模拟节点上运行此函数:
fixBounds() {
const graphNodes = this.forceSimulation.nodes();
graphNodes.forEach((node) => {
if (node.x - this.nodeRadius < 0) {
node.x = this.nodeRadius;
node.vx = 0;
}
if (node.y - this.nodeRadius < 0) {
node.y = this.nodeRadius;
node.vy = 0;
}
if (this.width && node.x + this.nodeRadius > this.width) {
node.x = this.width - this.nodeRadius;
node.vx = 0;
}
if (this.height && node.y + this.nodeRadius > this.height) {
node.y = this.height - this.nodeRadius;
node.yx = 0;
}
})
}
这将导致节点卡在边界上,节点之间的力将使它们远离边界。