說我有以下的HTML如何使用XPath和DOM替換php中的節點/元素?
$html = '
<div class="website">
<div>
<div id="old_div">
<p>some text</p>
<p>some text</p>
<p>some text</p>
<p>some text</p>
<div class="a class">
<p>some text</p>
<p>some text</p>
</div>
</div>
<div id="another_div"></div>
</div>
</div>
';
而且我想用下面的更換#old_div
:
$replacement = '<div id="new_div">this is new</div>';
舉的最終結果:
$html = '
<div class="website">
<div>
<div id="new_div">this is new</div>
<div id="another_div"></div>
</div>
</div>
';
有一個簡單的剪切和粘貼功能使用PHP做到這一點?
最後的工作代碼感謝所有戈登的幫助:
<?php
$html = <<< HTML
<div class="website">
<div>
<div id="old_div">
<p>some text</p>
<p>some text</p>
<p>some text</p>
<p>some text</p>
<div class="a class">
<p>some text</p>
<p>some text</p>
</div>
</div>
<div id="another_div"></div>
</div>
</div>
HTML;
$dom = new DOMDocument;
$dom->loadXml($html); // use loadHTML if it's invalid XHTML
//create replacement
$replacement = $dom->createDocumentFragment();
$replacement ->appendXML('<div id="new_div">this is new</div>');
//make replacement
$xp = new DOMXPath($dom);
$oldNode = $xp->query('//div[@id="old_div"]')->item(0);
$oldNode->parentNode->replaceChild($replacement , $oldNode);
//save html output
$new_html = $dom->saveXml($dom->documentElement);
echo $new_html;
?>
可能重複的[使用DOMXPath替換節點,同時保持其位置...](http://stackoverflow.com/questions/640815/using-domxpath-to-replace-a-node-while-維護其位置) – Gordon 2011-01-06 11:36:40
您可以使用** XPath來選擇**目標節點以**替換DOM方法**。反映標題。 – 2011-01-06 12:11:42