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 l开发者_高级运维ine. 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 :-)
Try
$result = preg_replace('%(\r?\n(?! \* ?@))?^(/\*\*\r?\n \* | \*/| \* ?)%m', ' ', $subject);
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
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