I want to add an additional form element in the configuration form for core/modules/views/src/Plugin/Block/ViewsExposedFilterBlock.php by adding

public function buildConfigurationForm(array $form, FormStateInterface $form_state) {
  $form = parent::buildConfigurationForm($form, $form_state);
  $form['my_new_form_element'] = [...];

  return $form;

If I hack core and add that code in the file, it does what I want. But of course I don't want to hack core. So I'm trying to figure out how to make it so that my custom module's block class is used instead of ViewsExposedFilterBlock.php.

So how do I alert Views to use my class, ExposedFilterImagesBlock.php?

Information that I've found that might help you help me:

ViewsExposedFilterBlock.php has deriver = "Drupal\views\Plugin\Derivative\ViewsExposedFilterBlock" in its annotation. So I'm assuming the deriver plays an important role...just one that is unclear to me.

Thanks in advance for any help you can provide!

  • 2
    You don't alert Views. All Views are scanned for configured exposed blocks and a block plugin definition is derived for each one found. You can try to replace the plugin class in hook_block_alter() and check if the derived blocks also have your custom class, see drupal.stackexchange.com/questions/264349/…
    – 4uk4
    Jun 26, 2019 at 20:35

1 Answer 1


Thanks to @4k4 (here) and @nikathone and @dorf (via Slack), I was able to arrive at this, placed in exposed_filter_images.module (all caps should be replaced with a machine name):

function exposed_filter_images_block_alter(&$plugins) {
  if (isset($plugins['views_exposed_filter_block:VIEW_NAME-DISPLAY_NAME'])) {
    $plugins['views_exposed_filter_block:VIEW_NAME-DISPLAY_NAME']['class'] = \Drupal\exposed_filter_images\Plugin\Block\ExposedFilterImagesBlock::class;

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