How to write a recursive regex that matches nested parentheses?

Question

I am trying to write a regexp which matches nested parentheses, e.g.:

\"(((text(text))))(text()()text)(casual(characters(#$%^^&&#^%#@!&**&#^*         


        

Answer 1

The following code uses my Parser class (it's under CC-BY 3.0), it works on UTF-8 (thanks to my UTF8 class).

The way it works is by using a recursive function to iterate over the string. It will call itself each time it finds a (. It will also detect missmatched pairs when it reaches the end of the string without finding the corresponding ).

Also, this code takes a $callback parameter you can use to process each piece it finds. The callback recieves two parameters: 1) the string, and 2) the level (0 = deepest). Whatever the callback returns will be replaced in the contents of the string (this changes are visible at callback of higher level).

Note: the code does not includes type checks.

Non-recursive part:

function ParseParenthesis(/*string*/ $string, /*function*/ $callback)
{
    //Create a new parser object
    $parser = new Parser($string);
    //Call the recursive part
    $result = ParseParenthesisFragment($parser, $callback);
    if ($result['close'])
    {
        return $result['contents'];
    }
    else
    {
        //UNEXPECTED END OF STRING
        // throw new Exception('UNEXPECTED END OF STRING');
        return false;
    }
}

Recursive part:

function ParseParenthesisFragment(/*parser*/ $parser, /*function*/ $callback)
{
    $contents = '';
    $level = 0;
    while(true)
    {
        $parenthesis = array('(', ')');
        // Jump to the first/next "(" or ")"
        $new = $parser->ConsumeUntil($parenthesis);
        $parser->Flush(); //<- Flush is just an optimization
        // Append what we got so far
        $contents .= $new;
        // Read the "(" or ")"
        $element = $parser->Consume($parenthesis);
        if ($element === '(') //If we found "("
        {
            //OPEN
            $result = ParseParenthesisFragment($parser, $callback);
            if ($result['close'])
            {
                // It was closed, all ok
                // Update the level of this iteration
                $newLevel = $result['level'] + 1;
                if ($newLevel > $level)
                {
                    $level = $newLevel;
                }
                // Call the callback
                $new = call_user_func
                (
                    $callback,
                    $result['contents'],
                    $level
                );
                // Append what we got
                $contents .= $new;
            }
            else
            {
                //UNEXPECTED END OF STRING
                // Don't call the callback for missmatched parenthesis
                // just append and return
                return array
                (
                    'close' => false,
                    'contents' => $contents.$result['contents']
                );
            }
        }
        else if ($element == ')') //If we found a ")"
        {
            //CLOSE
            return array
            (
                'close' => true,
                'contents' => $contents,
                'level' => $level
            );
        }
        else if ($result['status'] === null)
        {
            //END OF STRING
            return array
            (
                'close' => false,
                'contents' => $contents
            );
        }
    }
}

Answer 2

This pattern works:

$pattern = '~ \( (?: [^()]+ | (?R) )*+ \) ~x';

The content inside parenthesis is simply describe:

"all that is not parenthesis OR recursion (= other parenthesis)" x 0 or more times

If you want to catch all substrings inside parenthesis, you must put this pattern inside a lookahead to obtain all overlapping results:

$pattern = '~(?= ( \( (?: [^()]+ | (?1) )*+ \) ) )~x';
preg_match_all($pattern, $subject, $matches);
print_r($matches[1]);

Note that I have added a capturing group and I have replaced (?R) by (?1):

(?R) -> refers to the whole pattern (You can write (?0) too)
(?1) -> refers to the first capturing group

What is this lookahead trick?

A subpattern inside a lookahead (or a lookbehind) doesn't match anything, it's only an assertion (a test). Thus, it allows to check the same substring several times.

If you display the whole pattern results (print_r($matches[0]);), you will see that all results are empty strings. The only way to obtain the substrings found by the subpattern inside the lookahead is to enclose the subpattern in a capturing group.

Note: the recursive subpattern can be improved like this:

\( [^()]*+ (?: (?R) [^()]* )*+ \)

Answer 3

When I found this answer I wasn't able to figure out how to modify the pattern to work with my own delimiters which where { and }. So my approach was to make it more generic.

Here is a script to generate the regex pattern with your own variable left and right delimiters.

$delimiter_wrap  = '~';
$delimiter_left  = '{';/* put YOUR left delimiter here.  */
$delimiter_right = '}';/* put YOUR right delimiter here. */

$delimiter_left  = preg_quote( $delimiter_left,  $delimiter_wrap );
$delimiter_right = preg_quote( $delimiter_right, $delimiter_wrap );
$pattern         = $delimiter_wrap . $delimiter_left
                 . '((?:[^' . $delimiter_left . $delimiter_right . ']++|(?R))*)'
                 . $delimiter_right . $delimiter_wrap;

/* Now you can use the generated pattern. */
preg_match_all( $pattern, $subject, $matches );