preg_replace帶錨文本的href錨點

Question

如何將每個錨文本替換爲所有的錨點。我的代碼是

$body='<p>The man was <a href="http://www.example.com/video/">dancing like a little boy</a> while all kids were watching ... </p>'; 

我想要得到的結果是：

<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); 

和結果是：

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

Answer 1

試試這個

$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

您可以在這裏簡單地使用strip_tags()和htmlspecialchars()。

用strip_tags - 地帶HTML與字符串PHP標籤

用htmlspecialchars - 轉換特殊字符爲HTML實體

第1步：使用用strip_tags（）到去除除<p>標記之外的所有標記。

第2步：因爲我們需要與HTML標記一起獲得字符串，我們需要使用用htmlspecialchars（）。

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

當有已經是一個內置的PHP函數，我認爲這是更好和更緊湊的使用而不是使用preg_replace

Answer 3

可以使用此代碼：

正則表達式：/< 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); 

測試，並顯示例如匹配字：https://regex101.com/r/mgYjoB/1

Answer 4

沒有正則表達式.....

<?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]);