PHP, nested templates in preg_replace

Question

preg_replace("/\[b\](.*)\[\/b\]/Usi", "<strong>$1</strong>", "Some text here... [b][b]Hello, [b]PHP![/b][/b][/b] ... [b]and here[/b]");

returns

Some text here... <strong>[b]Hello, [b]PHP!</strong>[/b][/b] ... <strong>and here</strong>

But i need to replace all [b]...[/b] tags. Why this doesn't happen in my case?

Answer 1

Why use regex for this particular case? You could get away with a simple string replace every [b] to strong and every [/b] to the /strong.

Answer 2

Yes, a multi-pass approach is required if the elements are nested. This can be accomplished in one of two ways; matching from the inside out or from the outside in. Here are two tested scripts with fully commented regexes which illustrate each technique:

1. Replace from the inside out:

<?php // test.php Rev:20121016_0900
$re = '% # Match innermost [b]...[/b] structure.
    \[b\]              # Literal start tag.
    (                  # $1: Element contents.
      # Use Friedls "Unrolling-the-Loop" technique:
      #   Begin: {normal* (special normal*)*} construct.
      [^[]*            # {normal*} Zero or more non-"[".
      (?:              # Begin {(special normal*)*}.
        \[             # {special} Tag open literal char,
        (?!/?b\])      # but only if NOT [b] or [/b].
        [^[]*          # More {normal*}.
      )*               # Finish {(special normal*)*}.
    )                  # $1: Element contents.
    \[/b\]             # Literal end tag.
    %x';
printf("Replace matching tags from the inside out:\n");
$text = file_get_contents('testdata.txt');
$i=0; // Keep track of iteration number.
printf("i[%d]=%s", $i++, $text);
while(preg_match($re, $text)){
    $text = preg_replace($re, '<strong>$1</strong>', $text);
    printf("i[%d]=%s", $i++, $text);
}
?>

Output:

'''
Replace matching tags from the inside out:
i[0]=Some text here... [b][b]Hello, [b]PHP![/b][/b][/b] ... [b]and here[/b]
i[1]=Some text here... [b][b]Hello, <strong>PHP!</strong>[/b][/b] ... <strong>and here</strong>
i[2]=Some text here... [b]<strong>Hello, <strong>PHP!</strong></strong>[/b] ... <strong>and here</strong>
i[3]=Some text here... <strong><strong>Hello, <strong>PHP!</strong></strong></strong> ... <strong>and here</strong>
'''

2. Replace from the outside in:

<?php // test.php Rev:20121016_0901
$re = '% # Match outermost [b]...[/b] structure.
    \[b\]              # Literal start tag.
    (                  # $1: Element contents.
      (?:              # Zero or more contents alternatives.
        [^[]*          # Either non-[b]...[/b] stuff...
        (?:            # Begin {(special normal*)*}.
          \[           # {special} Tag open literal char,
          (?!/?b\])    # but only if NOT [b] or [/b].
          [^[]*        # More {normal*}.
        )*             # Finish {(special normal*)*}.
      | (?R)           # Or a nested [b]...[/b] structure.
      )*               # Zero or more contents alternatives.
    )                  # $1: Element contents.
    \[/b\]             # Literal end tag.
    %x';
printf("Replace matching tags from the outside in:\n");
$text = file_get_contents('testdata.txt');
$i=0; // Keep track of iteration number.
printf("i[%d]=%s", $i++, $text);
while(preg_match($re, $text)){
    $text = preg_replace($re, '<strong>$1</strong>', $text);
    printf("i[%d]=%s", $i++, $text);
}
?>

Output:

'''
Replace matching tags from the outside in:
i[0]=Some text here... [b][b]Hello, [b]PHP![/b][/b][/b] ... [b]and here[/b]
i[1]=Some text here... <strong>[b]Hello, [b]PHP![/b][/b]</strong> ... <strong>and here</strong>
i[2]=Some text here... <strong><strong>Hello, [b]PHP![/b]</strong></strong> ... <strong>and here</strong>
i[3]=Some text here... <strong><strong>Hello, <strong>PHP!</strong></strong></strong> ... <strong>and here</strong>
'''

Note the (?R) recursive expression used in the second approach.

Answer 3

The reason it doesn't work: You catch the first [b], then move on to the next [/b], and leave anything in between unchanged. Ie, you change the outer [b] tags, but not the ones nested inside.

Your comment to @meza suggests you want to replace the pseudo tags in pairs, or else leave them untouched. The best way to do this is to use multiple passes, like this

$markup = "Some text here... [b][b]Hello, [b]PHP![/b][/b][/b] ... [b]and here[/b]";
$count = 0;
do {
    $markup = preg_replace("/\[b\](.*?)\[\/b\]/usi", "<strong>$1</strong>", $markup, -1, $count );
} while ( $count > 0 );

print $markup;

I'm not even sure if you can do it in a one-line regex, but even if you could, it would be rather complex and therefore hard to maintain.

Answer 4

edit your modifiers Usi and replace it with sim.

EDIT:

Give this a shot:

<?php

   function matchReplaceAll($reg, $replace, $str)
    {
        while (preg_match($reg, $str))
        {
            $str = preg_replace($reg, $replace, $str);
        }
        return $str;
    }



$str="Some text here... [b][b]Hello, [b]PHP![/b][/b][/b] ... and here";
$str=matchReplaceAll('/\[b\](.*?)\[\/b\]/sim', '<strong>$1</strong>', $str);
echo $str;