使用强制定向图,当目标和源是同一节点时,如何获得实际显示的链接。所以基本上只是一个很好的小循环,表明存在这样的边缘。
我已经使用或试图使用过两个D3示例:
答案 0 :(得分:15)
诀窍是将自链接绘制为带有弧的路径。我花了一些时间来摆弄arc parameter syntax让事情变得有效,关键似乎是电弧无法在同一点开始和结束。以下是在每次更新时绘制边缘的相关代码。
function tick() {
link.attr("d", function(d) {
var x1 = d.source.x,
y1 = d.source.y,
x2 = d.target.x,
y2 = d.target.y,
dx = x2 - x1,
dy = y2 - y1,
dr = Math.sqrt(dx * dx + dy * dy),
// Defaults for normal edge.
drx = dr,
dry = dr,
xRotation = 0, // degrees
largeArc = 0, // 1 or 0
sweep = 1; // 1 or 0
// Self edge.
if ( x1 === x2 && y1 === y2 ) {
// Fiddle with this angle to get loop oriented.
xRotation = -45;
// Needs to be 1.
largeArc = 1;
// Change sweep to change orientation of loop.
//sweep = 0;
// Make drx and dry different to get an ellipse
// instead of a circle.
drx = 30;
dry = 20;
// For whatever reason the arc collapses to a point if the beginning
// and ending points of the arc are the same, so kludge it.
x2 = x2 + 1;
y2 = y2 + 1;
}
return "M" + x1 + "," + y1 + "A" + drx + "," + dry + " " + xRotation + "," + largeArc + "," + sweep + " " + x2 + "," + y2;
});
这里有a jsfiddle演示了整个内容和截图: