Drupal 6：views_get_view_result

Question

我需要检索视图的结果并在之后将其编码为 json。通过这样做，它向我显示“数组”而不是内容。你能告诉我哪里错了吗？非常感谢！

function web_service_expositions($viewname, $display_id = NULL, $args = NULL) {
  $view = views_get_view('expositions_now');
  if (is_object($view)) {
    if (is_array($args)) {
      $view->set_arguments($args);
    }
    if (is_string($display_id)) {
      $view->set_display($display_id);
    }
    else {
      $view->init_display();
    }
    $view->pre_execute();
    $view->execute();
/*  print "<pre> $viewname: $display_id";
    print_r(get_class_methods($view));  */
    return $view->result;
  }
  else {
    return t('View %viewname not found.', array('%viewname' => $viewname));
  }
}

Answer 1

尝试替换

返回$view->结果；

与

drupal_json_output($view->result);
drupal_exit();

https://api.drupal.org/api/drupal/includes%21common.inc/function/drupal_json_output/7