4

I'm trying to load a form into a div using a javascript call. I can get the form in the div, but nothing works (autocomplete fields, wysiwyg fields, custom submit handler).

I know I must be missing something, but I can't figure out what it is.

What I have now to get the form:

$form = \Drupal::formBuilder()->getForm('Drupal\my_module\Form\MyForm');
$renderable = [
   '#theme' => 'my_form',
   '#form' => $form,
];
$rendered = Drupal::service('renderer')->renderPlain($renderable);

$response = new AjaxResponse();
$response->setData($rendered);
return $response;

I've also tried renderRoot instead of renderPlain, but to no avail.

The twig file looks like this (very simple):

<div id="form-container">
    {% include 'header.html.twig' with {
        type : 'manual'
    } %}
    <div class="content">
        {{ form }}
    </div>
</div>

Any help will be greatly appreciated.

1 Answer 1

4

You have to add ajax commands to the ajax response, using the unrendered render array:

$response->addCommand(new ReplaceCommand('#form-container', $renderable));

BTW you can only return a Drupal ajax response if you have triggered an ajax request via ajax.js. Outside of form submits you usually do this by adding the class use-ajax to a link. For custom js code see How to Trigger Existing, Non-Form Ajax from JavaScript Event

2
  • Thanks, unfortunately this was not a route I could take, so now I have pre-loaded the form on the page (using twig) and display/hide the form when it is needed. Sep 29, 2020 at 7:26
  • OK, you don't need Ajax if you can pre-load static content. But a lot of use cases are more dynamic.
    – 4uk4
    Sep 29, 2020 at 7:41

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