preg_replace带锚文本的href锚点

Question

如何将每个锚文本替换为所有的锚点。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>'; 

我想要得到的结果是：

<p>The man was dancing like a little boy while all kids were watching ... </p> 

我用：

$body= preg_replace('#<a href="https?://(?:.+\.)?ok.co.*?>.*?</a>#i', '$1', $body); 

和结果是：

<p>The man was while all kids were watching ... </p> 

Answer 1

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); 

Answer 2

您可以在这里简单地使用strip_tags()和htmlspecialchars()。

用strip_tags - 地带HTML与字符串PHP标签

用htmlspecialchars - 转换特殊字符为HTML实体

第1步：使用用strip_tags（）到去除除<p>标记之外的所有标记。

第2步：因为我们需要与HTML标记一起获得字符串，我们需要使用用htmlspecialchars（）。

echo htmlspecialchars(strip_tags($body, '<p>')); 

当有已经是一个内置的PHP函数，我认为这是更好和更紧凑的使用而不是使用preg_replace

Answer 3

可以使用此代码：

正则表达式：/< 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); 

测试，并显示例如匹配字：https://regex101.com/r/mgYjoB/1

Answer 4

没有正则表达式.....

<?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]);