I have a page with a simple view. Above the view there is a menu link that when clicked loads a node form with a required file field. The form is fetched and loaded using ajax_commands into a div on the current page.

The problem is when I submit the form without providing a file for the required file field (intentionally triggering error), form_set_error redirects the user to the menu path originally called to produce the form (sitename-ajax/new-profile), which displays an unrendered ajax response. How do I get the validation response and return it to the target div on the same page – preventing a page redirect? Code pasted below is simplified for reference. Thanks!

In my hook_menu():

$items['sitename-ajax/new-profile'] = array(   'title' => t('New Profile'),   'type' => MENU_CALLBACK,   'page callback' => 'sitename_ajax_new_profile',   'access arguments' => array('access content'), ); 

Page callback:

function sitename_ajax_new_profile() {   $form = sitename_forms_prepare_form('ps_profile');    $output = drupal_render($form);    $commands = array();   $commands[] = ajax_command_html('#new-form-target', $output);    $page = array('#type' => 'ajax', '#commands' => $commands);    ajax_deliver($page); } 

Form helper function:

function sitename_forms_prepare_form($type) {   global $user;   module_load_include('inc', 'node', 'node.pages');    $node = (object) array(     'uid' => $user->uid,     'name' => (isset($user->name) ? $user->name : ''),     'type' => $type,     'language' => LANGUAGE_NONE   );    $output = drupal_get_form($type . '_node_form', $node);   return $output; }