正则表达式剥离phpdoc多行注释 [英] Regex to strip phpdoc multiline comment

问题描述

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剥离@token数据以外的所有数据，因此结果将是：

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

我现在有这个：

$text = preg_replace('/\r?\n *\* */', ' ', $text);

部分完成此工作：仅删除每行前面的*。谁能帮助我，让它也删除/ **和最后的斜杠/？

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!

PS：例如，如果commentlbock中包含

P.S: If, for instance, the commentlbock would contain something like

/**
 * @foo Here's some slashes for ya: / and \
 */

然后显然@foo之后的斜杠可能不会被删除。结果必须是：

Then obviously the slashes after @foo may not be stripped. The reult would have to be:

@foo Here's some slashes for ya: / and \

我希望那里有一个正则表达式专家：-）

I hope there's a regex guru out there :-)

推荐答案

尝试

$result = preg_replace('%(\r?\n(?! \* ?@))?^(/\*\*\r?\n \* | \*/| \* ?)%m', ' ', $subject);

它将在每行的开头插入一个额外的空格，因此您可能希望去除前导空白

It will insert an extra space at the start of each line, so you might want to strip leading whitespace in a second step.

说明：

（\r？\ n（?! \ *？@））？：如果可能，请匹配换行符，除非其后跟随 * @

(\r?\n(?! \* ?@))?: If possible, match a newline unless it's followed by * @

^ ：声明以下匹配项从行的开头开始

^: Assert that the following match starts at the beginning of the line

（：要么匹配

/ \ * \ * \r ？\n \ * ： / **< newline> *

| 或

\ * / ： * /

| ：或

\ *？： * ，可选后跟另一个空格

\* ?: *, optionally followed by another space

）：轮换序列的结尾

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