如何有选择地删除DOM文档的子树中的节点？

Question

我正在使用DOMDocument解析一个html文档并从中获取一些数据。以下是DOM的子树的结构

<div id="tab1"> 
    <div class="some class name"></div> 
    <div class="some other class name">arbitrary data and nodes</div> 
    <p> lot of paragraphs to follow </p> 
    <p> paragraphs </p> 
    <p> paragraphs </p> 
    <p> paragraphs </p> 
    <p> paragraphs </p> 
    <br /> 
    <br /> 
    <br /> 
    <br /> 
    <br /> 
    <table /> 
    <table /> 
    <table /> 
    <table /> 
</div> 

我不想要tab1的前两个孩子。我使用下面的PHP代码

<?php 
$urlArray = file('sitemap.txt'); 
$dataSet = array(); 
foreach($urlArray as $url){ 
    $scrapedData = file_get_contents('./scraped-site/'.trim($url)); 
    $doc = new DOMDocument(); 
    @$doc->loadHTML($scrapedData); 
    $domXpathDoc = new DOMXPath($doc); 
    $results = ''; 
    $xpathArray = array(
     'info'=>'//*[@id="tabs1"]', 
    ); 
    $set = array(); 
    foreach($xpathArray as $field => $xpath){ 
     $domNodeList = $domXpathDoc->query($xpath); 
     foreach($domNodeList as $node){ 
      foreach ($node->childNodes as $child) { 
       $set[] = $child->ownerDocument->saveXML($child); 
      } 
     } 
    } 
    $dataSet[] = $set; 
} 

给出的代码给了我所有的孩子我怎么能选择性地避免任何节点？

Answer 1

[EDIT2：我试过如下（我学:)）答案。这是为我工作：

"//*[@id='tabs1'][name() != 'div']" 

基本上它告诉XPath来忽略名为“格”的所有元素。您可以忽略这样一个以上的元素：

"//*[@id='tabs1'][name() != 'div' and name() != 'foo']" 

只显示前两个元素后，将工作是这样的：

"//*[@id='tabs1'][position()>2]"