如何根据当前节点属性 return 添加节点

Question

我们有这样的结构：

<class>
  <class-intro>
    <indication>Some content</indication>
  </class-intro>

  <article>
    <indication>Special content</indication>
  </article>

  <article includeclass="no">
    <indication>Different content</indication>
  </article>
</class>

我正在尝试 select 这些 XQuery/XPath 在每篇文章的基础上：

indication | node()[not(@includeclass) | @includeclass='yes']/ancestor::class/class-intro/indication

注意 - 我正在使用 PHP 的 http://php.net/manual/en/class.domxpath.php

// $xpath is a DOMXPath for the above document
$articles = $xpath->query("//article");

$indications = array();
foreach ($articles as $article) {
  $indications[] = $xpath->query(
    "indication | node()[not(@includeclass) | @includeclass='yes']/ancestor::class/class-intro/indication",
    $article
  );
}

var_dump($indications);

我希望得到：

array(
  0 => array(
    0 => "Some content",
    1 => "Special content",
  ),
  1 => array(
    0 => "Different content",
  ),
);

但我得到：

array(
  0 => array(
    0 => "Some content",
    1 => "Special content",
  ),
  1 => array(
    0 => "Some content",
    1 => "Different content",
  ),
);

Answer 1

问题是因为在此上下文中，对于每个 node()，not(@includeclass) 始终计算为 true，因为 article 的子元素的 none 具有属性 includeclass.

您应该使用 self 轴来引用当前上下文节点，即使用 self::node() 而不是 node()，因为 includeclass 属性属于当前上下文元素 article, 不是到子节点:

self::node()[not(@includeclass) or @includeclass='yes']/ancestor::class/.....

如何根据当前节点属性 return 添加节点

How to return additional node depending on current node attribute

php

xml

xpath

domxpath