Question

我有这个片段，显示网站中属于特定词汇表的所有分类列表。而不是打印整个列表，我如何只打印属于我实际加载的节点的术语？我有一个Drupal 7安装。

这是我打印节点ID的方式：<?php print $node->nid;?>

<?php 
$vid = 11; //vocabulary id

$query = "SELECT tid, name, count
FROM (
SELECT td.tid AS tid, name, COUNT(td.tid) AS count
FROM taxonomy_term_data AS td
JOIN taxonomy_index AS tn
  ON td.tid = tn.tid
JOIN node AS n
  ON n.nid = tn.nid
WHERE td.vid = ". $vid ."
  AND n.status = 1
GROUP BY td.tid
ORDER BY count DESC
) AS t
ORDER BY name ASC";
$result = db_query($query);
foreach($result as $term) {
    if ($term->count > 0) {
      echo l($term->name, "taxonomy/term/$term->tid").' ('.$term->count.')'.'<br/>';
    }
}
 ?>

Answer 1

我建议不要为此运行额外的查询。此信息应该在$ node对象中可用。

只需打印它[print_r（$ node）]，看看分类对象名称究竟是什么（$ node-＆gt; taxonomy）＆amp;分类信息的结构和方式用它来显示节点页面或节点预告片上的类别。

在其他页面上，您可以使用node_load来加载节点，然后执行相同的操作。

Answer 2

在这种情况下，

sumoand 的答案更加优化，但是对于一些在这里练习的SQL来说，这是你想象的确切解决方案：

<?php 
$vid = 11; //vocabulary id

$query = "SELECT tid, name, count
FROM (
SELECT td.tid AS tid, name, COUNT(td.tid) AS count
FROM taxonomy_term_data AS td
JOIN taxonomy_index AS tn
  ON td.tid = tn.tid
JOIN node AS n
  ON n.nid = tn.nid
WHERE td.vid = ". $vid ."
  AND n.status = 1
  AND n.nid = ".$node->id."
GROUP BY td.tid
ORDER BY count DESC
) AS t
ORDER BY name ASC";
$result = db_query($query);
foreach($result as $term) {
    if ($term->count > 0) {
      echo l($term->name, "taxonomy/term/$term->tid").' ('.$term->count.')'.'<br/>';
    }
}
?>

如何仅打印属于当前帖子的类别？此代码段显示了网站中的所有类别

2 个答案: