我如何使用PHP删除空文本节点的标签？ [英] How can I use php to remove tags with empty text node?

问题描述

如何使用php删除空文本节点的标签？

例如，

< div class =box>< / div> 删除

< ; a href =＃>< / a> 删除

< p>< a href =＃>< / a>< / p> 删除

< span style =...>< / span> remove

但是我想用这样的文本节点来保存标签

< a href =＃>链接< / a> 保持

编辑：

我想删除这样的东西，

 < p>< strong>< a href =http://xx.org.uk/dartmoor-arts>< / a> ;< /强>< / p为H. 
< p>< strong>< a href =http://xx.org.uk/depw>< / a>< / strong>< / p> 
< p>< strong>< a href =http://xx.org.uk/devon-guild-of-craftsmen>< / a>< / strong>< / p为H. 
  

我测试了两个正则表达式，

  $ content = preg_replace（'！<（。*？）[^>]> \s *< / \\ 1>！'，''，$ content ）; 
 $ content = preg_replace（'％<（。*？）[^>] *> \\\s *< / \\1>％'，''，$ content） ; 
  

但他们留下这样的东西，

 < p为H.;<强>< /强>< / p为H. 
< p>< strong>< / strong>< / p> 
< p>< strong>< / strong>< / p> 
  

解决方案

一种方法可能是：

  $ dom = new DOMDocument（）; 
 $ dom-> loadHtml（
'< p>< strong>< a href =http://xx.org.uk/dartmoor-arts>测试< / a> ;< / strong>>< / p> 
< p>< strong>< a href =http://xx.org.uk/depw>< / a>< / strong>< / p> 
< p>< strong>< a href =http://xx.org.uk/devon-guild-of-craftsmen>< / a> ;< / strong>< / p>'
）; 
 
 $ xpath = new DOMXPath（$ dom）; $（）
 $ b while（（$ nodeList = $ xpath-> query（'// * [not（text（））and not（node（））]'））&& $ nodeList- >长度> 0）{
 foreach（$ nodeList as $ node）{
 $ node-> parentNode-> removeChild（$ node）; 
} 
} 
 
 echo $ dom-> saveHtml（）; 
  

可能您必须稍微改变一下您的需求。

How can I use php to remove tags with empty text node?

For instance,

<div class="box"></div> remove

<a href="#"></a> remove

<p><a href="#"></a></p> remove

<span style="..."></span> remove

But I want to keep the tag with text node like this,

<a href="#">link</a> keep

Edit:

I want to remove something messy like this too,

<p><strong><a href="http://xx.org.uk/dartmoor-arts"></a></strong></p>
<p><strong><a href="http://xx.org.uk/depw"></a></strong></p>
<p><strong><a href="http://xx.org.uk/devon-guild-of-craftsmen"></a></strong></p>

I tested both regex below,

$content = preg_replace('!<(.*?)[^>]*>\s*</\1>!','',$content);
$content = preg_replace('%<(.*?)[^>]*>\\s*</\\1>%', '', $content);

But they leave something like this,

<p><strong></strong></p>
<p><strong></strong></p>
<p><strong></strong></p>

解决方案

One way could be:

$dom = new DOMDocument();
$dom->loadHtml(
    '<p><strong><a href="http://xx.org.uk/dartmoor-arts">test</a></strong></p>
    <p><strong><a href="http://xx.org.uk/depw"></a></strong></p>
    <p><strong><a href="http://xx.org.uk/devon-guild-of-craftsmen"></a></strong></p>'
);

$xpath = new DOMXPath($dom);

while(($nodeList = $xpath->query('//*[not(text()) and not(node())]')) && $nodeList->length > 0) {
    foreach ($nodeList as $node) {
        $node->parentNode->removeChild($node);
    }
}

echo $dom->saveHtml();

Probably you'll have to change that a bit for your needs.

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