Question

我要删除图像源（src）中的任何数量的“ ../”。

我有一个正则表达式，它删除了“ ../”，并且如果源代码中没有“ ../”，它也可以工作。

<?php

$help_path = 'http:xxx.com/wp-content/help/';
$contents = <<<MYVAR
<img src="Links/WAS_PIC_ControlBox-1-2-3.jpg" alt="WAS-Betjeningsboks-1-2-3" 
style="border: none; margin-left: 20px; margin-right: 0px; margin-top: 0px; margin-bottom: 0px;" border="0">

<img src="../Links/WAS_PIC_ControlBox-1-2-3.jpg" alt="WAS-Betjeningsboks-1-2-3" 
style="border: none; margin-left: 20px; margin-right: 0px; margin-top: 0px; margin-bottom: 0px;" border="0">

<img src="../../Links/WAS_PIC_ControlBox-1-2-3.jpg" alt="WAS-Betjeningsboks-1-2-3" 
style="border: none; margin-left: 20px; margin-right: 0px; margin-top: 0px; margin-bottom: 0px;" border="0">

<script type="text/javascript" src="ehlpdhtm.js"></script>

<script type="text/javascript" src="../ehlpdhtm.js"></script>

<script type="text/javascript" src="../../ehlpdhtm.js"></script>

MYVAR;

$find = array(
    '#<script\s+type="(.*?)"\s+src="(?:\.{2}/)?(.*?)">(.*?)</script>#is',
    '#<img\s+src="(?:\.{2}/)?(.*?)"\s+alt="(.*?)"([^>]*?)/?>#i'
);

$replace = array(
    '<script type="${1}" src="' . $help_path . '${2}"></script>',
    '<img src="' . $help_path . '${1}" alt="${2}"${3} />'
);

$preg_rep = preg_replace($find, $replace, $contents);
print_r($preg_rep);

问题是，如果src中有多个“ ../”，我的正则表达式将输出“ ../”。无论源中有多少个“ ../”（“../../../../”），它都应该始终像src =“ Links / xxx.jpg”。

<img src="http:xxx.com/wp-content/help/Links/WAS_PIC_ControlBox-1-2-3.jpg" alt="WAS-Betjeningsboks-1-2-3" 
style="border: none; margin-left: 20px; margin-right: 0px; margin-top: 0px; margin-bottom: 0px;" border="0" />

<img src="http:xxx.com/wp-content/help/Links/WAS_PIC_ControlBox-1-2-3.jpg" alt="WAS-Betjeningsboks-1-2-3" 
style="border: none; margin-left: 20px; margin-right: 0px; margin-top: 0px; margin-bottom: 0px;" border="0" />

<img src="http:xxx.com/wp-content/help/../Links/WAS_PIC_ControlBox-1-2-3.jpg" alt="WAS-Betjeningsboks-1-2-3" 
style="border: none; margin-left: 20px; margin-right: 0px; margin-top: 0px; margin-bottom: 0px;" border="0" />

<script type="text/javascript" src="http:xxx.com/wp-content/help/ehlpdhtm.js"></script>

<script type="text/javascript" src="http:xxx.com/wp-content/help/ehlpdhtm.js"></script>

<script type="text/javascript" src="http:xxx.com/wp-content/help/../ehlpdhtm.js"></script>

Answer 1

您可以使用preg_replace删除所有../

$contents = str_replace('../', '', $contents);

然后，您可以应用正则表达式

$contents = str_replace('../', '', $contents);
$find = array(
  '#<script\s+type="(.*?)"\s+src="(?:\.{2}/)?(.*?)">(.*?)</script>#is',
  '#<img\s+src="(?:\.{2}/)?(.*?)"\s+alt="(.*?)"([^>]*?)/?>#i'
);
$replace = array(
  '<script type="${1}" src="' . $help_path . '${2}"></script>',
  '<img src="' . $help_path . '${1}" alt="${2}"${3} />'
);

$preg_rep = preg_replace($find, $replace, $contents);

如何用'../'匹配和替换img源

1 个答案: