Reset node form after AJAX submit

Question

I have a node form in a block, that is submitted using AJAX.

I've created my own callback, because I'm adding the created node to the page if the submission was successful.

Now, I want to reset the node form to the default values. Any ideas on how to accomplish that?

Update: I've managed to replace the form, by looking at the way node_add() works.

$new_node = (object) array(
  'uid' => $user->uid,
  'name' => (isset($user->name) ? $user->name : ''),
  'type' => 'node_type',
  'language' => LANGUAGE_NONE,
);

And then, I'm replacing the entire form with the return value from drupal_render(drupal_get_form('node_form', $new_node));

The form gets replaced as it should be, apart from some ids being different. However, when I submit the new form, I'll get a javascript alert window, with the entire contents of a new page. I can see from the HTML, that it's the page for the node that has been created. I suppose the AJAX callback gets redirected to the new node, but I don't know why.

Any clues?

Answer 1

Got it!

The main problem was that the new form was loaded from a cached version of the previous submitted form.

This is what I've got in my AJAX callback:

$new_form_state = array();
$new_form_state['build_info']['args'][] = (object) array(
  'uid' => $user->uid,
  'name' => (isset($user->name) ? $user->name : ''),
  'type' => 'node_type',
  'language' => LANGUAGE_NONE,
);
$new_form_state['no_redirect'] = TRUE;
$new_form_state['input'] = array();
$new_form = drupal_build_form('node_type_node_form', $new_form_state);
$commands[] = ajax_command_replace('#node-type-node-form', drupal_render($new_form));

The $new_form_state['no_redirect'] = TRUE; prevents the redirect, and the empty array in $new_form_state['input'] prevents the form from being loaded from a cached version.

@googletorp, thanks for the help!

Answer 2

If you are are using Drupal 7 you can unset form_state['values'] and rebuild the form which should work (untested).

An alternative solution would be to issue ajax commands and update the form with the default values fetched from the form definition.

Answer 3

I have the same task. Node add form must be posted via AJAX. After node is successfully created - form must be ready to add next node (i.e. all form fields must be empty). If form is not validated - it must be shown to user again to allow him fix errors.

This is my solution

function MY_MODULE_form_CONTENT_TYPE_node_form_alter(&$form, &$form_state, $form_id) {

  $form['actions']['submit']['#ajax'] = array(
    'wrapper' => $form_state['node']->type . '-node-form',
    'callback' => 'MY_MODULE_node_add_callback',
    'method' => 'replace',
    'effect' => 'fade',
  );

  // We can have several forms on one page.
  // This id must be unique, because ajax behavior is attached using it.
  $form['actions']['submit']['#id'] = 'edit-submit-' . $form_state['node']->type . '-ajax';
}

function MY_MODULE_node_add_callback($form, &$form_state) {

  $node = $form_state['node'];

  if (!empty($node->nid)) {

    // Node is created successfully.
    // Return the same from but with empty fields.
    $new_node = (object) array(
      'uid' => $node->uid,
      'name' => $node->name,
      'type' => $node->type,
      'language' => $node->language,
    );

    $new_form_state = array();
    $new_form_state['build_info']['args'] = array($new_node);
    $new_form_state['values'] = array();
    $new_form_state['method'] = $form_state['method'];

    form_load_include($new_form_state, 'inc', 'node', 'node.pages');

    return drupal_rebuild_form($node->type . '_node_form', $new_form_state);
  }
  else {

    // Node is not created because of validation fail.
    // Return current form so user can continue editing and fix errors.
    return $form;
  }
}

Answer 4

Use drupal_rebuild_form() function. http://api.drupal.org/api/drupal/includes%21form.inc/function/drupal_rebuild_form/7