I am trying to create a web component that is usable in a form element specifically, that has a name
, and a value
. I recognize that I can create a web component that extends
an HTMLInputElement
:
<input is="very-extended">
but I am trying to create an entirely new element.
When creating a regular web component, you can create it from the prototype of a regular HTMLElement
(HTMLElement.prototype
). Which leads me to assume that I can create a different element with the prototype of an HTMLInputElement
(HTMLInputElement.prototype
). You actually use that prototype when extending
the API of input elements, so why can't I use that prototype to create an entirely new element that works in a form?
If you look at the shadow dom of a regular input field:
you can see that there is a div inside of there. I understand that this HTMLInputElement
has methods and attributes, getters/setters, etc. So why is it that when I attempt to create my component, it fails to be part of the name, value pairs found in the form?
Here is an example of how I am trying to create this web component:
Please note that his should be tested in a browser that supports web components.
(function() {
var iconDoc = (document._currentScript || document.currentScript).ownerDocument;
var objectPrototype = Object.create(HTMLInputElement.prototype);
Object.defineProperty(objectPrototype, 'name', {
writable: true
});
Object.defineProperty(objectPrototype, 'value', {
writable: true
});
objectPrototype.createdCallback = function() {
var shadow = this.createShadowRoot();
var template = iconDoc.querySelector('#test');
shadow.appendChild(template.content.cloneNode(true));
};
document.registerElement('custom-input', {
prototype: objectPrototype
});
})();
console.log(
$('form').serialize()
)
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<template id="test">
<div>This is a special input</div>
</template>
<form>
<input name="regular" value="input">
<custom-input name="foo" value="bar"></custom-input>
</form>
Why is the name, value pair not found among the form, and how can I create a custom form element?
You can create your <custom-input>
custom element that will be interpreted by a form
, just by adding inside your template
a hidden input
element with the
name
and value
pair your want.
<template>
<input type="hidden" name="foo" value="defaultVal">
</template>
The default value
(and name
) can by updated by your custom element internal logic.
This hidden input
must not be inserted inside a Shadow DOM to be detected by the container form
.
(function() {
var iconDoc = (document._currentScript || document.currentScript).ownerDocument;
var objectPrototype = Object.create(HTMLInputElement.prototype);
objectPrototype.createdCallback = function() {
//var shadow = this.createShadowRoot();
var template = iconDoc.querySelector('#test');
this.appendChild(template.content.cloneNode(true));
};
document.registerElement('custom-input', {
prototype: objectPrototype
});
})();
console.log(
$('form').serialize()
)
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<template id="test">
<input type="hidden" name="foo" value="bar">
</template>
<form>
<input name="regular" value="input">
<custom-input name="foo" value="bar"></custom-input>
</form>
KevBot,
You seem to think that the element includes itself in the form. That is not the case. It is the form that searches its children elements by tag name, to decide which elements it should include. It will simply ignore those with unknown tag names.
Your custom-input
name is not among the elements the form searches. Therefore, it is not included on the form. It doesn't matter the custom element's prototype. That's why it works if you use is
, since then the tag name is maintained.
Of course, you may implement your own custom-form
that behaves differently, if you want.
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