Php DOM和Xpath -替换节点，但保留旧节点的子节点 - Php DOM and Xpath - Replace node but keep children of old node

Php DOM and Xpath - Replace node but keep children of old node

考虑以下html:

<html>
    <title>Xyz</title>
    <body>
        <div>
            <div class='mycls'>
                <div>1 Books</div>
                <div>2 Papers</div>
                <div>3 Pencils</div>
            </div>
        </div>
    <body>
</html>
$dom = new DOMDocument();
$dom->loadHTML([loaded html of remote url through curl]);
$xpath = new DOMXPath($dom);
$nodes = $xpath->query('html/body/div[@class="mycls"]');

直到这里工作良好，我需要替换节点得到以下:

<body>
        <div>
            <span>
                <div>1 Books</div>
                <div>2 Papers</div>
                <div>3 Pencils</div>
            </span>
        </div>
    <body>

下面的内容应该可以为您工作:

$dom = new DOMDocument();
$dom->loadHTML($html);
$xpath = new DOMXPath($dom);
$oldNode = $xpath->query('//div[@class="mycls"]')->item(0);
$span = $dom->createElement('span');
if ($oldNode->hasChildNodes()) {
    $children = [];
    foreach ($oldNode->childNodes as $child) {
        $children[] = $child;
    }
    foreach ($children as $child) {
        $span->appendChild($child->parentNode->removeChild($child));
    }
}
$oldNode->parentNode->replaceChild($span, $oldNode);
echo htmlspecialchars($dom->saveHTML());

演示:http://codepad.viper - 7. - com/wntrr5

请注意，在演示中，我还修复了你的HTML，这是完全损坏的:-)

如果你的demo确实是你从cURL调用中得到的HTML，并且你不能改变它(无法控制它)，你可以这样做:

$libxmlErrors = libxml_use_internal_errors(true); // at the start

和

libxml_use_internal_errors($libxmlErrors); // at the end

防止错误弹出