D3力布局：根据节点半径沿链接移动箭头

Question

在此D3 example中：

节点具有可变半径。我设法移动节点标签，以便它们始终紧挨着它们的圈子。

但是，如何移动箭头？ （你可以看到，对于微软，Apple等，他们几乎被圈子所覆盖）

相关问题：

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Answer 1

要根据节点半径调整箭头的位置，可以调整路径的终点。更新了jsFiddle

function linkArc(d) {
    var targetX = d.target.x - d.target.started,
      targetY = d.target.y - d.target.started,
      dx = targetX - d.source.x,
      dy = targetY - d.source.y,
      dr = (d.straight == 0)?Math.sqrt(dx * dx + dy * dy):0;
  return "M" + d.source.x + "," + d.source.y +
       " L " + targetX + "," + targetY;
}

Answer 2

我在线搜索，答案都没有，所以我自己做了：

以下是代码：

   //arrows
svg.append("defs").selectAll("marker")
    .data(["suit", "licensing", "resolved"])
    .enter().append("marker")
    .attr("id", function(d) { return d; })
    .attr("viewBox", "0 -5 10 10")
    .attr("refX", 9)
    .attr("refY", 0)
    .attr("markerWidth", 10)
    .attr("markerHeight", 10)
    .attr("orient", "auto")
    .append("path")
    .attr("d", "M0,-5L10,0L0,5 L10,0 L0, -5")
    .style("stroke", "#4679BD")
    .style("opacity", "0.6"); 

  //Create all the line svgs but without locations yet
var link = svg.selectAll(".link")
   .data(forceData.links)
   .enter().append("line")
   .attr("class", "link")
   .style("marker-end", "url(#suit)");

//Set up the force layout
var force = d3.layout.force()
    .nodes(forceData.nodes)
    .links(forceData.links)
    .charge(-120)
    .linkDistance(200)
    .size([width, height])
    .on("tick", tick)
    .start();

function tick(){
    link.attr("x1", function (d) { return d.source.x; })
        .attr("y1", function (d) { return d.source.y; })
        .attr("x2", function (d) { 
            return calculateX(d.target.x, d.target.y, d.source.x, d.source.y, d.target.radius); 
        })
        .attr("y2", function (d) { 
            return calculateY(d.target.x, d.target.y, d.source.x, d.source.y, d.target.radius);
        });

    d3.selectAll("circle")
        .attr("cx", function (d) { return d.x; })
        .attr("cy", function (d) { return d.y; });

    d3.select("#forcelayoutGraph").selectAll("text")
        .attr("x", function (d) { return d.x; })
        .attr("y", function (d) { return d.y; });
}
function calculateX(tx, ty, sx, sy, radius){
    if(tx == sx) return tx;                 //if the target x == source x, no need to change the target x.
    var xLength = Math.abs(tx - sx);    //calculate the difference of x
    var yLength = Math.abs(ty - sy);    //calculate the difference of y
    //calculate the ratio using the trigonometric function
    var ratio = radius / Math.sqrt(xLength * xLength + yLength * yLength);
    if(tx > sx)  return tx - xLength * ratio;    //if target x > source x return target x - radius
    if(tx < sx) return  tx + xLength * ratio;    //if target x < source x return target x + radius
}
function calculateY(tx, ty, sx, sy, radius){
    if(ty == sy) return ty;                 //if the target y == source y, no need to change the target y.
    var xLength = Math.abs(tx - sx);    //calculate the difference of x
    var yLength = Math.abs(ty - sy);    //calculate the difference of y
    //calculate the ratio using the trigonometric function
    var ratio = radius / Math.sqrt(xLength * xLength + yLength * yLength);
    if(ty > sy) return ty - yLength * ratio;   //if target y > source y return target x - radius
    if(ty < sy) return ty + yLength * ratio;   //if target y > source y return target x - radius
}