How to change programmatically exposed filters in views

Question

I have a page view with exposed filters. One of the filters is term vocabulary, and that's dropdown list, user can select only one category (term).

I want to set programmatically one term from dropdown list when I came from specific URL. I tried with views_query_alter where $category is tid.

 $set_exposed_input  = array();
 $set_exposed_input['field_category_tid'] = $category;

 if ($set_exposed_input) {
    $view->set_exposed_input($set_exposed_input);
 }

Answer 1

For drupal 7 you can use

$view->exposed_input[$key] = $value;

This is

$view->exposed_input['title'] = 'My title is a cat';

For a full example you can try this:

$name = 'name_of_view';
$display = 'page_1';
$alter = array('exposed' =>array('title' => 'title_value'));

$view = views_get_view($name);
$view->init_display($display);
$view->preview=TRUE;
$view->is_cacheable = FALSE;


if(isset($alter['exposed'])){
  foreach($alter['exposed'] as $key => $valor){

     $view->exposed_input[$key] = $valor;

  }
}

$view->pre_execute();
$output = $view->display_handler->preview();
$view->post_execute();

print_r($view->result);

Other way is using the views set_option function

$filters = $view->display_handler->get_option('filters');

foreach($alter['exposed'] as $key => $valor){
   $filters[$key]['value'] = $valor;
   $view->display_handler->set_option('filters', $filters);
}

Answer 2

with Drupal 7.54 when you want to process the exposed filter programmatically, this is how you do it when you have a multiple valued field. The solution to see how to pass exposed filter is to follow how the URL is made.

Here: view_name?alias[1]=1

$view = views_get_view('view_name');
$view->init_display('display_id');

$filters = array('alias' => array('1'));
$view->set_exposed_input($filters);

$view->execute();

You can execute this from wherever you want. Preferably from the module file.

Answer 3

Set exposed filter value like this:

$view->exposed_input['field_category_tid'] = $category;