ElRudi
ElRudi

Reputation: 2234

d3 create object without appending

I'm using d3 for graphing, and I'm trying to create an svg object, to add to the DOM later.

I used to have

var svg = d3.select(el).append("svg");
var graph = svg.append("g")
...etc...

and for reasons I won't go into, I wanted to create the svg element before appending it to the DOM.

So I did

var svg = d3.select(document.createElementNS(d3.ns.prefix.svg, 'svg'))
var graph = svg.append("g")
...etc...

, which works, and while debugging, I can see that svg is a 1-element array, with the children nicely attached.

The problem comes at the append step:

d3.select(el).append(svg);

There, I get an error Error: Failed to execute 'createElementNS' on 'Document': The qualified name provided ('[object SVGSVGElement]') contains the invalid name-start character '['. I've taken a look here: How to create "svg" object without appending it? but it seems that's exactly how they suggest it.

Any idea why this is so? I've tried appending svg[0], but no luck either. It seems append() only takes strings as an argument.


edit: the d3 reference at https://github.com/mbostock/d3/wiki/Selections#append states

selection.append(name) ... ... ... "The name may be specified either as a constant string or as a function that returns the DOM element to append."

Consequently I've tried

d3.select(el).append(function(){return svg;});

but that fails with a Error: Failed to execute 'appendChild' on 'Node': The new child element is null.

Upvotes: 10

Views: 8215

Answers (2)

jshanley
jshanley

Reputation: 9138

You might be better off using the regular element.appendChild function for this.

You can get a reference to your parent element using d3.select(el).node() then, you can call .appendChild on that, passing in svg.node() as the element to append.

So, all together:

d3.select(el).node().appendChild(svg.node());

Upvotes: 4

Stephen Thomas
Stephen Thomas

Reputation: 14063

If svg is a selection, svg.node() returns the DOM element, e.g.:

d3.select(el).append(function(){return svg.node();});

(Mind you, I'm not certain that svg in your case is a true selection, but you could give it a try.)

Upvotes: 16

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