Python Regex中评论代码

Question

我想在大多数文件的开头注释掉代码中匹配开源许可证类型。但是，对于期望的字符串（例如较低通用公共许可证）跨越两行的情况，我遇到了困难。例如，查看许可证下面的代码。

* Copyright (c) Codice Foundation 
* <p/> 
* This is free software: you can redistribute it and/or modify it under the terms of the GNU Lesser 
* General Public License as published by the Free Software Foundation, either version 3 of the 
* License, or any later version. 
* <p/> 
* This program is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without 
* even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU 
* Lesser General Public License for more details. A copy of the GNU Lesser General Public License 
* is distributed along with this program and can be found at 
* <http://www.gnu.org/licenses/lgpl.html>. 
*/ 

使用正则表达式的回溯是不可能的，因为在注释代码空间未知数量以及在不同的语言不同的注释字符。我目前正则表达式的例子包括如下：

self._cr_license_re['GNU']       = re.compile('\sGNU\D') 
self._cr_license_re['MIT License']     = re.compile('MIT License|Licensed MIT|\sMIT\D') 
self._cr_license_re['OpenSceneGraph Public License'] = re.compile('OpenSceneGraph Public License', re.IGNORECASE) 
self._cr_license_re['Artistic License']    = re.compile('Artistic License', re.IGNORECASE) 
self._cr_license_re['LGPL']       = re.compile('\sLGPL\s|Lesser General Public License', re.IGNORECASE) 
self._cr_license_re['BSD']       = re.compile('\sBSD\D') 
self._cr_license_re['Unspecified OS']     = re.compile('free of charge', re.IGNORECASE) 
self._cr_license_re['GPL']       = re.compile('\sGPL\D|(?<!Lesser)\sGeneral Public License', re.IGNORECASE) 
self._cr_license_re['Apache License']     = re.compile('Apache License', re.IGNORECASE) 
self._cr_license_re['Creative Commons']    = re.compile('\sCC\D') 

我欢迎就如何解决Python中使用正则表达式这个问题的任何建议。

Answer 1

你可以使用this regex，并用空格

\s*\*\s*\/? 

这种替换应该把多行注释在同一行，那么你就可以找到它的许可证。