正则表达式去除 phpdoc 多行注释
我有这个：
/**
* @file
* API for loading and interacting with modules.
* More explaination here.
*
* @author Reveller <me@localhost>
* @version 19:05 28-12-2008
*/
我正在寻找一个正则表达式来删除除 @token 数据之外的所有数据，所以结果是：
@file API for loading and interacting with modules. More explaination here.
@author Reveller <me@localhost>
@version 19:05 28-12-2008
我现在有这个：
$text = preg_replace('/\r?\n *\* */', ' ', $text);
它部分完成了工作：它只删除了每行前面的 * 。谁能帮助我，让它也去掉 /** 和最后的斜杠 /？任何帮助将不胜感激！
PS：例如，如果 commentlbock 包含类似的内容
/**
* @foo Here's some slashes for ya: / and \
*/
，那么显然 @foo 之后的斜杠可能不会被删除。结果一定是：
@foo Here's some slashes for ya: / and \
我希望有一个正则表达式大师:-)
I have this:
/**
* @file
* API for loading and interacting with modules.
* More explaination here.
*
* @author Reveller <me@localhost>
* @version 19:05 28-12-2008
*/
I'm looking for a regex to strip all but the @token data, so the result would be:
@file API for loading and interacting with modules. More explaination here.
@author Reveller <me@localhost>
@version 19:05 28-12-2008
I now have this:
$text = preg_replace('/\r?\n *\* */', ' ', $text);
It does the job partially: it only removes the * in front of each line. Who could help me so it also strips /** and the final slash /? Any help would be greatly appreciated!
P.S: If, for instance, the commentlbock would contain something like
/**
* @foo Here's some slashes for ya: / and \
*/
Then obviously the slashes after @foo may not be stripped. The reult would have to be:
@foo Here's some slashes for ya: / and \
I hope there's a regex guru out there :-)
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Try
It 会在每行的开头插入一个额外的空格，因此您可能需要在第二步中删除前导空格。
说明：
(\r?\n(?! \* ?@))?：如果可能，匹配换行符，除非后面跟着* @^< /code>: 断言以下匹配从行的开头开始(: 匹配/\*\*\r?\n \*:/**<换行> *|或\*/:*/|:或\* ?：*，可选地后跟另一个空格)：交替序列结束Try
It will insert an extra space at the start of each line, so you might want to strip leading whitespace in a second step.
Explanation:
(\r?\n(?! \* ?@))?: If possible, match a newline unless it's followed by* @^: Assert that the following match starts at the beginning of the line(: Either match/\*\*\r?\n \*:/**<newline> *|or\*/:*/|: or\* ?:*, optionally followed by another space): End of alternation sequence