PHP PCRE允许在字符串中嵌套模式（递归）

Question

I got a string like 1(8()3(6()7())9()3())2(4())3()1(0()3()) which is representing a tree. A bracket appears, if we go one level deeper. Numbers on the same level are neighbours.

Now want to add nodes, for example I want to add a 5 to every path where we have a 1 on the first and a 3 on the second level, so I want to put a 5() after every 3( which is inside of a 1( . So 5() has to be added 3 times and the result should be 1(8()3(5()6()7())9()3(5()))2(4())3()1(0()3(5()))

The Problem is, that I don't get the code working with the PCRE recursion. If I match a tree representation string without and fixed paths like 1( and 3( it works, but when I generate a regex with those fixed patterns, it doesn't work.

This is my code:

<?php
header('content-type: text/plain;utf-8');

$node = [1, 3, 5];
$path = '1(8()3(6()7())9()3())2(4())3()1(0()3())';

echo $path.'
';

$nes = '\((((?>[^()]+)|(?R))*)\)';
$nes = '('.$nes.')*';

echo preg_match('/'.$nes.'/x', $path) ? 'matches' : 'matches not';
echo '
';

// creates a regex with the fixed path structure, but allows nested elements in between
// in this example something like: /^anyNestedElementsHere 1( anyNestedElementsHere 3( anyNestedElementsHere ))/
$re = $nes;
for ($i = 0; $i < count($node)-1; $i++) {
    $re .= $node[$i].'\(';
    if ($i != count($node)-2)
        $re .= $nes;
}
$re = '/^('.$re.')/x';

echo str_replace($nes, '   '.$nes.'   ', $re).'
';
echo preg_match($re, $path) ? 'matches' : 'matches not';
echo '
';
// append 5()
echo preg_replace($re, '${1}'.$node[count($node)-1].'()', $path);
?>

And this is the output, where you can see how the generated regex looks like:

1(8()3(6()7())9()3())2(4())3()1(0()3())
matches
/^(   (\((((?>[^()]+)|(?R))*)\))*   1\(   (\((((?>[^()]+)|(?R))*)\))*   3\()/x
matches not
1(8()3(6()7())9()3())2(4())3()1(0()3())

I hope you understand my problem and hope you can tell me, where my error is.

Thanks a lot!

Answer 1

Solution

Regex

The following regex matches nested brackets recursively, finding an opening 1( on the first level, and an opening 3( on the second level (as a direct child). It also attempts successive matches, either on the same level or going down the respective levels to find another match.

~
(?(?=\A)  # IF: First match attempt (if at start of string)   - -

  # we are on 1st level => find next "1("

  (?<balanced_brackets>
    # consumes balanced brackets recursively where there is no match
    [^()]*+
    \(  (?&balanced_brackets)*?  \)
  )*?

  # match "1(" => enter level 2
  1\(

|         # ELSE: Successive matches  - - - - - - - - - - - - - -

  \G    # Start at end of last match (level 3)

  # Go down to level 2 - match ")"
  (?&balanced_brackets)*?
  \)

  # or go back to level 1 - matching another ")"
  (?>
    (?&balanced_brackets)*?
    \)

    # and enter level 2 again
    (?&balanced_brackets)*?
    1\(
  )*?
)                                      # - - - - - - - - - - - -

# we are on level 2 => consume balanced brackets and match "3("
(?&balanced_brackets)*?
3\K\(  # also reset the start of the match
~x

Replacement

(5()

Text

Input:
1(8()3(6()7())9()3())2(4())3()1(0()3())

Output:
1(8()3(5()6()7())9()3(5()))2(4())3()1(0()3(5()))
       ^^^            ^^^                  ^^^
       [1]            [2]                  [3]

regex101 demo

---

How it works

We start by using a conditional subpattern to distinguish between:

• the first match attempt (from level 1) and
• the successive attempts (starting at level 3, anchored with the \\G assertion ).

(?(?=\A)  # IF followed by start of string
    # This is the first attempt
|         # ELSE
    # This is another attempt
    \G    # and we'll anchor it to the end of last match
)

For the first match , we'll consume all nested brackets that don't match 1( , in order to get the cursor to a position in the first level where it could find a successful match.

• This is a well-known recursive pattern to match nested constructs. If you're unfamiliar with it, please refer to Recursion and Subroutines .

(?<balanced_brackets>        # ANY NUMBER OF BALANCED BRACKETS
  [^()]*+                    # match any characters 
  \(                         # opening bracket
    (?&balanced_brackets)*?  #  with nested bracket (recursively)
  \)                         # closing bracket in the main level
)*?                          # Repeated any times (lazy)

Notice this is a named group that we will use as a subroutine call many times in the pattern to consume unwanted balanced brackets, as (?&balanced_brackets)*? .

Next levels . Now, to enter level 2, we need to match:

1\(

And finally, we'll consume any balanced brackets until we find the opening of the 3rd level:

(?&balanced_brackets)*?
3\(

That's it. We've just matched our first occurrence, so we can insert the replacement text in that position.

Next match . For the successive match attempts, we can either:

• go down to level 2 matching a closing ) to find another occurrence of 3(
• go further down to level 1 matching 2 closing ) and, from there, match using the same strategy as we used for the first match.

This is achieved with the following subpattern:

\G                             # anchored to the end of last match (level 3)
(?&balanced_brackets)*?        # consume any balanced brackets
\)                             # go down to level 2
                               #
(?>                            # And optionally
  (?&balanced_brackets)*?      #   consume level 2 brackets
  \)                           #   to go down to level 1
  (?&balanced_brackets)*?      #   consume level 1 brackets
  1\(                          #   and go up to level 2 again
)*?                            # As many times as it needs to (lazy)

To conclude , we can match the opening of the 3rd level:

(?&balanced_brackets)*?
3\(

We'll also reset the start of match near the end of the pattern, with \\K , to only match the last opening bracket. Thus, we can simply replace with (5() , avoiding the use of backreferences.

---

PHP Code

We only need to call preg_replace() with the same values used above.

Ideone demo

---

Why did your regex fail?

Since you asked, the pattern is anchored to the start of string. It can only match the first occurrence.

/^(   (\((((?>[^()]+)|(?R))*)\))*   1\(   (\((((?>[^()]+)|(?R))*)\))*   3\()/x

Also, it doesn't match the first occurrence because the construct (?R) recurses the the whole pattern (trying to match ^ again). We could change (?R) to (?2) .

The main reason, though, is because it is not consuming the characters before any opening \\( . For example:

Input:
1(8()3(6()7())9()3())2(4())3()1(0()3())
  ^
  #this "8" can't be consumed with the pattern

There's also a behaviour that should be considered: PCRE treats recursion as atomic . So you have to make sure that the pattern will consume unwanted brackets as in the above example, but also avoid matching 1( or 3( in their respective levels.

Answer 2

I'd break down this problem into two smaller parts:

First, extract the 1 nodes, using the following regex:

(?(DEFINE)
  (?<tree>
    (?: \d+ \( (?&tree) \) )*
  )
)
\b 1 \( (?&tree) \)

Demo

Use preg_replace_callback for this. This will match 1(8()3(6()7())9()3()) and 1(0()3()) .

Next, it's just a matter of replacing 3( with 3(5() and you're done.

Example in PHP:

$path = '1(8()3(6()7())9()3())2(4())3()1(0()3())';

$path = preg_replace_callback('#
    (?(DEFINE)
      (?<tree>
        (?: \d+ \( (?&tree) \) )*
      )
    )
    \b 1 \( (?&tree) \)
#x', function($m) {
    return str_replace('3(', '3(5()', $m[0]);
}, $path);

The result is: 1(8()3(5()6()7())9()3(5()))2(4())3()1(0()3(5()))