Question

I’m using the better_exposed_filters module to create a set of exposed filters for a view. One of the filters is being displayed as a select field, and I would like the field to only display options that are actually associated with content in the database.

Currently, I am doing this using the hook_form_alter() method. For simplification, in the following example the field is called ‘foo’ and the content type with that field is called ‘bar’:

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

  // Get all the values of foo that matter
  $resource = db_query('select distinct field_foo_value from {content_type_bar}');
  $foo = array();
  while($row = db_fetch_object($resource)) {
    $foo[$row->field_foo_value] = $row->field_foo_value;
  }

  $form['foo']['#options'] = $foo;
}

This works great — the form displays only the options I want to display. Unfortunately, the view doesn’t actually display anything initially and I also get the following error message:

An illegal choice has been detected. Please contact the site administrator.

After I filter options with the form once, everything seems to work fine.
Does anyone know how I can solve this problem? I’m open to an entirely different way of weeding out filter options, if need be, or a way that I can figure out how to address that error.

Answer 1

I found a solution that works, but it’s somewhat hackish. I force the form to think that it’s validated, and it doesn’t complain anymore, with the following line at the bottom of the function:

$form['foo']['#validated'] = true;