Question

我正在使用DOMDocument来解析html文档并从中获取一些数据。以下是DOM

的子树结构

<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。我正在使用以下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如何有选择地避免任何节点？

Answer 1

[EDIT2：我尝试了下面的答案（我学到了:)）。这对我有用：

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

基本上它告诉xpath忽略所有名为'div'的元素。您可以忽略多个这样的元素：

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

只有在前两个元素之后显示元素才会这样：

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

如何有选择地删除DOM文档子树中的节点？

1 个答案: