用文本描述替換html鏈接標記

Question

我搜索了四周，似乎無法得到正確的答案。我想，以取代所有的鏈接在給定的HTML字符串是這樣的：

... <a href="link.php">Description Here</a> ... 
to: 
Description Here: link.php 

這是爲了讓非HTML電子郵件客戶端可以看到該鏈接，必須複製並粘貼到瀏覽器的能力。

Answer 1

你可以做這樣的事情：

$text = '<a href="link.php">Description Here</a>'; 

$replaced = preg_replace('/<a [^>]*href="(.*)"[^>]*>(.*)<\/a>/', '$2: $1', $text); 

Answer 2

PCRE拯救你：

$text = '<a href="link.php">Description Here</a>'; 

echo preg_replace("/\<a.*href=\"(.*)\".*\>(.*)\<\/a\>/Usi",'$2: $1',$text); 

Answer 3

echo preg_replace("/<a.*?href=\"(.*?)\".*?>(.*?)<\/a>/i", "$2: $1", $html); 

Answer 4

在純PHP從你可以一個通用的HTML文件：

<?php 
    //an example html string 
    $dataStr = '<ul id="sub-menu"> 
    <li class="parent item10"><a href="link1.php">Arts</a> 
     <ul> 
      <li class="item29"> 
       <a href="link2.php">Arts Show</a></li> 
      <li class="item29"> 
       <a href="link3.php">Another Arts Show</a></li> 
     </ul> 
    </li> 
    <li class="parent item15"><a href="link6.php">News</a> 
     <ul> 
      <li class="item18"> 
       <a href="link7.php">News Show</a></li> 
     </ul> 
    </li> 
</ul>'; 


    $startIndex = strpos($dataStr, "<a", 0); 
    $endIndex = strpos($dataStr, "</a", $startIndex); 

    while (1==1){ 
     if($startIndex == false || $endIndex == false){ 
      break; 
     } 

     $currentLine = substr($dataStr,$startIndex,($endIndex-$startIndex)+4); 

     $linkStartIndex = strpos($currentLine, "href=", 0)+6; 
     $linkEndIndex = strpos($currentLine, '"', $linkStartIndex); 
     $currentLink = substr($currentLine,$linkStartIndex,$linkEndIndex-$linkStartIndex); 

     $descriptionStartIndex = strpos($currentLine, ">", $linkEndIndex)+1; 
     $descriptionEndIndex = strpos($currentLine, "<", $descriptionStartIndex); 
     $currentDescription = substr($currentLine,$descriptionStartIndex,$descriptionEndIndex-$descriptionStartIndex); 

     echo ($currentDescription.": ".$currentLink."\n"); 

     $startIndex = strpos($dataStr, "<a", $endIndex+5); 
     $endIndex = strpos($dataStr, "</a", $startIndex); 

    } 
?>