Regex to insert line breaks after specific series of token - php

Question

I am trying to convert a multi-line string, to insert line breaks \n after two final closing parenthesis occur in series, ie: )) becomes ))\n.
There is also likely to be a ')' just prior to the '))', effectively creating ')))'.
These two or three parenthesis may or may not be "spread out" by indeterminate lengths of whitespace, eg )), ) ), ))), ) ) ), )) ), ) )) and so on.
I've tried the following:
//Example message
$message = '(item (name 286) (Index 31) (Image "item001") (class money coin) (code 4 110 0) (country 2) (plural 1) (buy 0))
(item(name 7904)(Index 7904) (specialty (Dex 10(defense 55)(hp 3500)(dodge 71) ))
(item(name 7905)(Index 7905)(country 2)
(level 80)(specialty(hp 3400) ) )
(item(name 7906)(Index 7906)(level 80) (specialty(Str 10)) ) ';
// Converts all lines into one line
$message = preg_replace("/[\r\n]*/","",$message);
// Replace '))' with '))\n' - doesn't work.
$message = preg_replace("/[)s+)]s*/","\n",$message);
$InititemLines = explode("\n", $message);
for ($line = 0; $line < count($InititemLines); $line++) {
echo "Line #<b>{$line}</b> : " . $InititemLines[$line] . "<br />\n";
}
To convert all lines into one, I used:
$message = preg_replace("/[\r\n]*/","",$message);
Then, to replace )) with ))\n, I tried the following (but it doesn't work):
$message = preg_replace("/[)s+)]s*/","))\n",$message);
I want the output to be like this:
Line #0: (item (name 286) (Index 31) (Image "item001") (class money coin) (code 4 11 0 0) (country 2) (plural 1) (buy 0))
Line #1: (item(name 7904)(Index 7904) (specialty (Dex 10)(defense 55)(hp 3500)(dodge 71) ))
Line #2: (item(name 7905)(Index 7905)(country 2)(level 80)(specialty(hp 3400) ) )
Line #3: (item(name 7906)(Index 7906)(level 80) (specialty(Str 10)) )

Answer 1

This will replace the "))" at the end of ALL lines, in the case of line ending in ))) or )):
$message = Preg_replace( "\)?(\s*\)\s*\))", "$1\n", $message );
This regex means
find an optional single closing parenthesis ')'. We escape it as ')' as the parenthesis has special meanings in regex. It's optional because of the trailing '?'.
followed by 0 or more space characters denoted as '\s'. 0 or more denoted by '*',
followed by another ')'
followed by another 0 or more spaces,
followed by another ')'
Then we surround \s*\)\s*\) with a pair of '(' and ')' meaning "group this section, so we can reference it later". We do this so we can replace it with ))\n.
And then a more elegant solution might be (depending on your requirements...), to subsequently also strip any excess remaining spaces from before every ')':
$message = preg_replace("(\)\s*)", "\)", $message);
This regex means
find an operning ')',
followed by 0 or more spaces
Grouped, so we can replace.
(In your example, I believe this will strip all the excess whitespace, while leaving the spaces in your strings alone).

Answer 2

Thank you got it working fine with
$message = Preg_replace("/(\s*\)\s*\)?\s*\))/", "$1\n", $message );