如何用每个锚文本替换所有锚。我的代码是
$body='<p>The man was <a href="http://www.example.com/video/">dancing like a little boy</a> while all kids were watching ... </p>';
我希望结果是:
<p>The man was dancing like a little boy while all kids were watching ... </p>
i used:
$body= preg_replace('#<a href="https?://(?:.+'.)?ok.co.*?>.*?</a>#i', '$1', $body);
,结果是:
<p>The man was while all kids were watching ... </p>
试试这个
$body='<p>The man was <a href="http://www.example.com/video/">dancing like a little boy</a> while all kids were watching ... </p>';
echo preg_replace('#<a.*?>([^>]*)</a>#i', '$1', $body);
无正则表达式.....
<?php
$d = new DOMDocument();
$d->loadHTML('<p>The man was <a href="http://www.example.com/video/">dancing like a little boy</a> while all kids were watching ... </p>');
$x = new DOMXPath($d);
foreach($x->query('//a') as $anchor){
$url = $anchor->getAttribute('href');
$domain = parse_url($url,PHP_URL_HOST);
if($domain == 'www.example.com'){
$anchor->parentNode->replaceChild(new DOMText($anchor->textContent),$anchor);
}
}
function get_inner_html( $node ) {
$innerHTML= '';
$children = $node->childNodes;
foreach ($children as $child) {
$innerHTML .= $child->ownerDocument->saveXML( $child );
}
return $innerHTML;
}
echo get_inner_html($x->query('//body')[0]);
可以使用以下代码:
regex:
/< a.*?>|<a.*?>|<'/a>/g
$body='<p>The man was <a href="http://www.example.com/video/">dancing like a little boy</a> while all kids were watching ... </p>';
echo preg_replace('/< a.*?>|<a.*?>|<'/a>/', ' ', $body);
test and show example match word: https://regex101.com/r/mgYjoB/1
您可以在这里简单地使用strip_tags()和htmlspecialchars()
strip_tags -从字符串中删除HTML和PHP标签
htmlspecialchars -将特殊字符转换为HTML实体
步骤1:使用strip_tags()剥离除<p>
标签外的所有标签。
第2步:由于我们需要获得字符串和HTML标记,我们需要使用htmlspecialchars()。
echo htmlspecialchars(strip_tags($body, '<p>'));
当已经有一个内置的PHP函数时,我认为使用它比使用preg_replace