How do I print an exposed filter block in a template file?

Question

I have a view with an exposed taxonomy filter. I configured it to be displayed as a separate block so that I can print it another twig template. I tried this.

/**
 * Implements hook_preprocess_block() for block.html.twig.
 */
 function themename_preprocess_block(array &$variables) {
   $variables['featured'] = views_embed_view('view_name', 'contextual_filter_block_machine_name');
 }

Then I used {{ featured }} in the block template, but it doesn't work.

Is there another way to do this?

Answer 1

I almost always use the twig_tweak module, there is 3 ways to render a block. You can see the doc here Twig tweak Block render, you should probably use

drush ev "print_r(array_keys(\Drupal::service('plugin.manager.block')->getDefinitions()));"

find your block in the list and then use

{{ drupal_block('your_block_id') }}

Answer 2

I used another approach to solve the problem. I overrode the views template to print the exposed filter where I wanted. You can get it as

{{ exposed}}

in the twig template.