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2013-11-21T09:09:09.660
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Question is quite dated, but since I don't see an answer and someone might face the same problem, here it is. The reason why simple replacement of diagonal with line is not working is because d3.svg.line and d3.svg.diagonal return different results: <ul> <li>d3.svg.diagonal returns function that accepts datum and its index and transforms it to path using projection. In other words diagonal.projection determines how the function will get points' coordinates from supplied datum.</li> <li>d3.svg.line returns function that accepts an array of points of the line and transforms it to path. Methods line.x and line.y determine how coordinates of the point retreived from the single element of supplied array</li> </ul> <a href="https://github.com/mbostock/d3/wiki/SVG-Shapes" rel="noreferrer">D3 SVG-Shapes reference</a> <a href="https://www.dashingd3js.com/svg-paths-and-d3js" rel="noreferrer">SVG Paths and D3.js</a> So you can not use result of the d3.svg.line directly in d3 selections (at least when you want to draw multiple lines). You need to wrap it in another function like this: <pre><code>var line = d3.svg.line() .x( function(point) { return point.lx; }) .y( function(point) { return point.ly; }); function lineData(d){ // i'm assuming here that supplied datum // is a link between 'source' and 'target' var points = [ {lx: d.source.x, ly: d.source.y}, {lx: d.target.x, ly: d.target.y} ]; return line(points); } // usage: var link= svg.selectAll("path") .data(links) .enter().append("path") .attr("d",lineData) .attr("class", ".link") .attr("stroke", "black") .attr("stroke-width", "2px") .attr("shape-rendering", "auto") .attr("fill", "none"); </code></pre> Here's working version of jsFiddle mobeets posted: <a href="http://jsfiddle.net/8xrU7/" rel="noreferrer">jsFiddle</a>
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