How do I programmatically render a view that has exposed filters?

Question

I can call a view with contextual filters without any problem, but when I add exposed filter value $view->setExposedInput to the call, I just does not work, the exposed value is not applied and being ignored. The filter 'price' field is set in the views filters options and i tried both exposing it and not exposing it to visitors without any luck. I call this view from my controller class and I need to pass the arguments to both filters and I do not need exposed form to be displayed to the user.

$view = Views::getView($view_id);
if (is_object($view)) {
  $filters = [
       'price' => 500, //my exposed filter value       
     ];      
  $view->setExposedInput($filters);
  $view->setArguments($args);
  $view->setDisplay($display_id);
  $view->preExecute();
  $view->execute();
  $view->buildRenderable($display_id, $args);
  $content = $view->render();
}

Answer 1

You can use this code in custom block can also set permission.

 use Drupal\Core\Form\FormState; 
 use Drupal\views\Views; 
   $form = []; 
   $view_id = 'blog_landing_page'; 
   $display_id = 'page_1'; 
   $view = Views::getView($view_id);     
   if ($view) { 
   $view->setDisplay($display_id); 
   $view->initHandlers(); 
 $form_state = (new FormState()) 
  ->setStorage([ 
   'view' => $view, 
   'display' => &$view->display_handler->display, 
   'rerender' => TRUE, 
   ]) 
   ->setMethod('get') 
   ->setAlwaysProcess() 
   ->disableRedirect(); 
   $form_state->set('rerender', NULL); 
 $form = \Drupal::formBuilder()->buildForm('\Drupal\views\Form\ViewsExposedForm', $form_state); 
    }