Drupal 7 - 将分类法插入节点对象

Question

我有一个成功创建新节点的脚本.但是在保存之前我无法设置分类法.

我相信Drupal 6我会使用这种方法.

$cat1_tid = taxonomy_get_term_by_name($data[$i]['cat1']);
$cat2_tid = taxonomy_get_term_by_name($data[$i]['cat2']);
$cat3_tid = taxonomy_get_term_by_name($data[$i]['cat3']);
$node->taxonomy = array($cat1_tid, $cat2_tid, $cat3_tid);

我想在Drupal 7中我会这样做(我的字段名是Catalog)

$node->taxonomy_catalog['und'][0] = array($term1Obj, $term2Obj);

taxonomy_get_term_by_name似乎没有返回要插入节点对象的正确对象.

如果有人可以发光,赞赏.

谢谢

编辑

解:

// Taxonomy
$categories = array($data[$i]['cat1'], $data[$i]['cat2'], $data[$i]['cat3']);
foreach ($categories as $key => $category) {
  if ($term = taxonomy_get_term_by_name($category)) {
    $terms_array = array_keys($term);
    $node->taxonomy_catalog[LANGUAGE_NONE][$key]['tid'] = $terms_array['0'];
  }   
} 

Answer 1

下面是我最近用来将“命令”节点导入站点的一些快速而肮脏的代码。在中间，foreach 循环负责根据需要创建和分配术语。

      $command = new stdClass;
      $command->language = LANGUAGE_NONE;
      $command->uid = 1;
      $command->type = 'drubnub';
      $command->title = $line['0'];
      $command->body[LANGUAGE_NONE]['0']['value'] = $line['1'];
      $command->url[LANGUAGE_NONE]['0']['value'] = trim($line['2']);
      $command->uses[LANGUAGE_NONE]['0']['value'] = $line['3'];
      $tags = explode(',', $line['4']);
      foreach ($tags as $key => $tag) {
        if ($term = taxonomy_get_term_by_name($tag)) {
          $terms_array = array_keys($term);
          $command->field_tags[LANGUAGE_NONE][$key]['tid'] = $terms_array['0'];
        } else {
          $term = new STDClass();
          $term->name = $tag;
          $term->vid = 1;
          if (!empty($term->name)) {
            $test = taxonomy_term_save($term);
            $term = taxonomy_get_term_by_name($tag);
            foreach($term as $term_id){
              $command->product_tags[LANGUAGE_NONE][$key]['tid'] = $term_id->tid;
          }
            $command->field_tags[LANGUAGE_NONE][$key]['tid'] = $tid;
          }
        }
      }
      node_save($command);