d3.js - D3 tree: lines instead of diagonal projection -

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i using d3.js create tree using this example.

this handles data have , produces desired outcome except 1 detail: don't want wiggly connector lines between nodes, want clean , simple line. can show me how produce that?

i've been looking @ api documentation d3.js, no success. understood, svg.line function should produce straight line given set of 2 pairs of coordinates (x,y). think need know is: given data, how create line given (cx,cy) of each pair of nodes in links array:

var margin = {top: 40, right: 40, bottom: 40, left: 40};  var width = 960 - margin.left - margin.right;   var height = 500 - margin.top - margin.bottom;  var tree = d3.layout.tree()     .size([height, width]);  var diagonal = d3.svg.diagonal()     .projection(function(d) { return [d.y, d.x]; });  var svg = d3.select("body").append("svg")     .attr("width", width + margin.left + margin.right)     .attr("height", height + margin.top + margin.bottom)     .append("g")     .attr("transform", "translate(" + margin.left + "," + margin.top + ")");  d3.csv("graph.csv", function(links) {     var nodesbyname = {};  links.foreach(function(link) { var parent = link.source = nodebyname(link.source),     child = link.target = nodebyname(link.target);     if (parent.children) parent.children.push(child);     else parent.children = [child]; });  var nodes = tree.nodes(links[0].source);  svg.selectall(".link")     .data(links) .enter().append("path")     .attr("class", "link")     .attr("d", diagonal);  svg.selectall(".node")     .data(nodes) .enter().append("circle")     .attr("class", "node")     .attr("r", 10)     .attr("cx", function(d) { return d.y; })     .attr("cy", function(d) { return d.x; });  function nodebyname(name) {     return nodesbyname[name] || (nodesbyname[name] = {name: name}); } });

actually figured out other example:

svg.selectall(".link")     .data(links) .enter().append("line")     .attr("class", "link")     .attr("x1", function(d) { return d.source.y; })     .attr("y1", function(d) { return d.source.x; })     .attr("x2", function(d) { return d.target.y; })     .attr("y2", function(d) { return d.target.x; });


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