PHP - preg_match/preg_replace problems - php

Question

I'm a little confused with preg_match and preg_replace. I have a very long content string (from a blog), and I want to find, separate and replace all [caption] tags. Possible tags can be:
[caption]test[/caption]
[caption align="center" caption="test" width="123"]<img src="...">[/caption]
[caption caption="test" align="center" width="123"]<img src="...">[/caption]
etc.
Here's the code I have (but I'm finding that's it not working the way I want it to...):
public function parse_captions($content) {
if(preg_match("/\[caption(.*) align=\"(.*)\" width=\"(.*)\" caption=\"(.*)\"\](.*)\[\/caption\]/", $content, $c)) {
$caption = $c[4];
$code = "<div>Test<p class='caption-text'>" . $caption . "</p></div>";
// Here, I'd like to ONLY replace what was found above (since there can be
// multiple instances
$content = preg_replace("/\[caption(.*) width=\"(.*)\" caption=\"(.*)\"\](.*)\[\/caption\]/", $code, $content);
}
return $content;
}

Answer 1

The goal is to ignore the content position. You can try this:
$subject = <<<'LOD'
[caption]test1[/caption]
[caption align="center" caption="test2" width="123"][/caption]
[caption caption="test3" align="center" width="123"][/caption]
LOD;
$pattern = <<<'LOD'
~
\[caption # begining of the tag
(?>[^]c]++|c(?!aption\b))* # followed by anything but c and ]
# or c not followed by "aption"
(?| # alternation group
caption="([^"]++)"[^]]*+] # the content is inside the begining tag
| # OR
]([^[]+) # outside
) # end of alternation group
\[/caption] # closing tag
~x
LOD;
$replacement = "<div>Test<p class='caption-text'>$1</p></div>";
echo htmlspecialchars(preg_replace($pattern, $replacement, $subject));
pattern (condensed version):
$pattern = '~\[caption(?>[^]c]++|c(?!aption\b))*(?|caption="([^"]++)"[^]]*+]|]([^[]++))\[/caption]~';
pattern explanation:
After the begining of the tag you could have content before ] or the caption attribute. This content is describe with:
(?> # atomic group
[^]c]++ # all characters that are not ] or c, 1 or more times
| # OR
c(?!aption\b) # c not followed by aption (to avoid the caption attribute)
)* # zero or more times
The alternation group (?| allow multiple capture groups with the same number:
(?|
# case: the target is in the caption attribute #
caption=" # (you can replace it by caption\s*+=\s*+")
([^"]++) # all that is not a " one or more times (capture group)
"
[^]]*+ # all that is not a ] zero or more times
| # OR
# case: the target is outside the opening tag #
] # square bracket close the opening tag
([^[]+) # all that is not a [ 1 or more times (capture group)
)
The two captures have now the same number #1
Note: if you are sure that each caption tags aren't on several lines, you can add the m modifier at the end of the pattern.
Note2: all quantifiers are possessive and i use atomic groups when it's possible for quick fails and better performances.

Answer 2

Hint (and not an answer, per se)
Your best method of action would be:
Match everything after caption.
preg_match("#\[caption(.*?)\]#", $q, $match)
Use an explode function for extracting values in $match[1], if any.
explode(' ', trim($match[1]))
Check the values in array returned, and use in your code accordingly.