Php:删除多个子标记(脚本)

Question

需要查找包含条件的单个<script> condition </script>。下面的示例包含4个标签，我需要匹配包含条件的第二个标签，并丢弃其他标签。从<script>开始。之前的条件可以是空格或换行符，然后像if (window.location.href == bar) { }和than这样的条件可以是空格或换行符，以及结束</script>。

<script>                  <!-- discard --->
other stuff
not to be found
</script>                 


<script>                  <!-- MATCH --->

if (window.location.href == bar) {         
do something                 
}

</script>                

<script>                 <!-- discard --->
other stuff
not to be found
</script>                

<script>                 <!-- discard --->
other stuff
not to be found
</script>

提前感谢

Answer 1

在这里，您应该使用DOMDocument而不是Regex来匹配标记及其所需内容。

<?php
ini_set('display_errors', 1);
$object= new DOMDocument();
$object->loadHTML('<html><body><script>                  <!-- discard --->
other stuff
not to be found
</script>                 

<script>                  <!-- MATCH --->

if (window.location.href == bar) {         
do something                 
}

</script>                

<script>                 <!-- discard --->
other stuff
not to be found
</script>                

<script>                 <!-- discard --->
other stuff
not to be found
</script></body></html>');
$tagsToRemove=array();
foreach($object->getElementsByTagName("script") as $element)
{
    if($element instanceof  DOMElement)
    {
        if(!preg_match("/if\s*\(/i", $element->nodeValue))
        {
            $tagsToRemove[]=$element;
        }
    }
}
foreach($tagsToRemove as $element)
{
    $element->parentNode->removeChild($element);
}
echo $object->saveHTML();