Is it possible for Prototype to enhance SVG components?

Question

Is it possible for Prototype to enhance SVG components?

Having some issues developing an interactive SVG map with Prototype. Extending inline SVG elements seems to be causing problems. Take a look at my code snippet below (path data removed for brevity):

<svg id="map" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" version="1.1" width="330" height="500" viewBox="-16037,-19651 28871,44234">
    <g id="state_outline">
        <path id="state" fill="white" stroke-width="200" d="..." />
        <path id="water" fill="#a0a0ff" d="..." />
    </g>
    <g id="counties">
        <path class="county" id="adams" d="..." />
        ...
    </g>
</svg>

<div id="nottamap"></div>       

<script type="text/javascript">
    console.log($('nottamap'));
    console.log($('nottamap').identify());
    console.log($('counties'));
    console.log($('counties').identify());
</script>

Upon execution, the output is:

<div id="nottamap">
nottamap
<g id="counties">
$("counties").identify is not a function

The issue here is that $() does not want to extend the element within the SVG structure. Is there a specific aspect of how Prototype interacts with XML elements that I might be overlooking, or perhaps a different approach I should consider?

javascript svg prototypejs

Answer 1

Answer №1

Prototype enhances elements using its Element.addMethods function. If you examine the source code, you can find this important section:

var elementPrototype = window.HTMLElement ? HTMLElement.prototype :
  Element.prototype;

if (F.ElementExtensions) {
  copy(Element.Methods, elementPrototype);
  copy(Element.Methods.Simulated, elementPrototype, true);
}

In this snippet, preference is given to the HTMLElement.prototype, which SVG elements don't inherit from. As a fallback, it resorts to using element.prototype in browsers like IE that lack support for SVG. You also have the option of directly modifying the source and replicating all actions to SVGElement.

While doing all this might seem like a lot of effort, there's a simpler workaround available. The static Element.* methods still function properly when used directly. So instead of using $('counties').identify(), you can use Element.identify('counties'). For example:

$$('.classname').invoke('hide');

You could opt for a more elegant functional approach like so:

$$('.classname').each(Element.hide);
// or replace Element.hide with Effect.fade or any other effect

The only downside is that you won't be able to utilize method chaining anymore.

Answer 2

Prototype enhances elements using its Element.addMethods function. If you examine the source code, you can find this important section:

var elementPrototype = window.HTMLElement ? HTMLElement.prototype :
  Element.prototype;

if (F.ElementExtensions) {
  copy(Element.Methods, elementPrototype);
  copy(Element.Methods.Simulated, elementPrototype, true);
}

In this snippet, preference is given to the HTMLElement.prototype, which SVG elements don't inherit from. As a fallback, it resorts to using element.prototype in browsers like IE that lack support for SVG. You also have the option of directly modifying the source and replicating all actions to SVGElement.

While doing all this might seem like a lot of effort, there's a simpler workaround available. The static Element.* methods still function properly when used directly. So instead of using $('counties').identify(), you can use Element.identify('counties'). For example:

$$('.classname').invoke('hide');

You could opt for a more elegant functional approach like so:

$$('.classname').each(Element.hide);
// or replace Element.hide with Effect.fade or any other effect

The only downside is that you won't be able to utilize method chaining anymore.

Answer 3

Answer №2

clockworkgeek's suggestion to modify the prototype source code had a positive impact on my project. In version 1.7 of the prototype, a simple adjustment in the relevant section achieved the desired outcome:

var elementPrototype = window.HTMLElement ? HTMLElement.prototype :
  Element.prototype;

if (F.ElementExtensions) {
  if (window.SVGElement) {
    copy(Element.Methods, SVGElement.prototype);
    copy(Element.Methods.Simulated, SVGElement.prototype, true);
  }
  copy(Element.Methods, elementPrototype);
  copy(Element.Methods.Simulated, elementPrototype, true);
}

For prototype version 1.7.1, the necessary changes are as follows:

var ELEMENT_PROTOTYPE = window.HTMLElement ? HTMLElement.prototype :
  Element.prototype;

if (F.ElementExtensions) {
  if (window.SVGElement) {
    mergeMethods(SVGElement.prototype, Element.Methods);
    mergeMethods(SVGElement.prototype, Element.Methods.Simulated, true);
  }
  mergeMethods(ELEMENT_PROTOTYPE, Element.Methods);
  mergeMethods(ELEMENT_PROTOTYPE, Element.Methods.Simulated, true);
}

The solution I implemented was sourced from this informative blog post

Answer 4

clockworkgeek's suggestion to modify the prototype source code had a positive impact on my project. In version 1.7 of the prototype, a simple adjustment in the relevant section achieved the desired outcome:

var elementPrototype = window.HTMLElement ? HTMLElement.prototype :
  Element.prototype;

if (F.ElementExtensions) {
  if (window.SVGElement) {
    copy(Element.Methods, SVGElement.prototype);
    copy(Element.Methods.Simulated, SVGElement.prototype, true);
  }
  copy(Element.Methods, elementPrototype);
  copy(Element.Methods.Simulated, elementPrototype, true);
}

For prototype version 1.7.1, the necessary changes are as follows:

var ELEMENT_PROTOTYPE = window.HTMLElement ? HTMLElement.prototype :
  Element.prototype;

if (F.ElementExtensions) {
  if (window.SVGElement) {
    mergeMethods(SVGElement.prototype, Element.Methods);
    mergeMethods(SVGElement.prototype, Element.Methods.Simulated, true);
  }
  mergeMethods(ELEMENT_PROTOTYPE, Element.Methods);
  mergeMethods(ELEMENT_PROTOTYPE, Element.Methods.Simulated, true);
}

The solution I implemented was sourced from this informative blog post

Is it possible for Prototype to enhance SVG components?

Answer №1

Answer №2

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