Is it possible to load page content without rendering html.tpl.php? I mean on default Drupal has the follwoing:

Html.tpl.php --> Page.tpl.php -->...(etc.)

my goal is to remove Html.tpl.php for a custom path (site://ajax-unesco)


page--ajax-unesco.tpl.php --> ... (etc.)

  • is ajax-unesco content type ? – Sandesh Yadav Jun 29 '13 at 13:47
  • keying in on the ajax portion of your path, if you are wanting Drupal to just output the $content of a page instead of all the various wrappings, in that menu callback, PRINT what you want instead of RETURNing it. – Jimajamma Jun 29 '13 at 13:49
  • It`s a view, i wanna get its contents by ajax call without additional data (like html.tpl contents). – Reza Shadman Jun 29 '13 at 13:50
  • @Jimajamma i didn't get it, currently i have a view with page display and a template file called "page--ajax-unesco.tpl.php (my view name) i'm rendering with the following code: <?php print render($page['content']); ?> – Reza Shadman Jun 29 '13 at 13:53
  • have you tried creating html--ajax-unesco.tpl.php in templates file ? you can create and keep it blank and test. – Sandesh Yadav Jun 29 '13 at 13:54

You could do something like this:

 * Implements hook_menu
function MY_MODULE_menu(){
  $items = array();
  $items['clean/view'] = array(
    'page callback'    => 'MY_MODULE_clean_view',
    'page arguments'   => array(2, 3),
    'access arguments' => array('access content'),
    'type'             => MENU_CALLBACK,
  return $items;

 * Menu callback for clean/view
 * Output view without wrapper page markup
function MY_MODULE_clean_view($name, $display = 'default'){
  // Note views_embed_view performs views access check
  print views_embed_view($name, $display);

You can then use the dynamic path /clean/view/view_name/display_id to render the View without the wrapper markup.


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