Only replace part of regex match

Question

I am looking to match all strings that have the combination _[[ or ]]_

That part I got down: (_\\[\\[)|(\\]\\]_)
Now comes the part where I need help though, how do I replace only the underscore in these instances?

In other words, the string: "_[[2, verb//substantiv//adjektiv]]_" would result in the string: "[[2, verb//substantiv//adjektiv]]"

Appreciate any help I can get.

Answer 1

The solution you can use here is to simply match the entire pattern and replace it with the same pattern without the enclosing underscores ( _ ).

I created the example here btw.

Example:

$str = 'My _[[string to parse]]_ with some _[[examples]]_';
$parsed = preg_replace('/_\[\[([^(\]\]_)]*?)\]\]_/', "[[$1]]", $str);
echo $parsed;

Output:

My [[string to parse]] with some [[examples]]

Regex explained:

• _\\[\\[ the starting point of the sequence you want to capture
• ([^((\\]\\]_))]*?) captures the contents of what is between the opening and closing sequence that is not the closing sequence itself
• \\]\\]_ the closing sequence

By matching the entire pattern and capturing the contents using a capture group you can replace the pattern entirely with a new substring that includes the contents from the matched pattern.

This is done in the second argument to preg_replace which is "[[$1]]"

$1 here stands for the captured group and contains its contents, which will be interpolated between two sets of square brackets.

Since the pattern also matches the underscores ( _ ) however, these are also removed but simply not replaced by anything in the second argument.

Answer 2

You could come up with:

$regex = '~
              _\[{2}  # look for an underscore and two open square brackets
              ([^]]+) # capture anything that is not a closing bracket
              \]{2}_  # followed by two closing square brackets and an underscore
          ~x';        # free space mode for this explanation
$string = "_[[2, verb//substantiv//adjektiv]]_";

# in the match replace [[(capture Group 1)]]
$new_string = preg_replace($regex, "[[$1]]", $string);
// new_string = [[2, verb//substantiv//adjektiv]]

See a demo on regex101.com as well as on ideone.com .

Answer 3

If you want to

match all strings that have the combination _[[ or ]]_

You can use this regex :

^(?=.*_\[\[).+|(?=.*\]\]_).+$

^               // start of the string
(?=.*_\[\[)     // if the string contains _[[
.+              // get the entire string (if the assert is correct)
|               // OR operands (if the assert is not correct, let's check the following)
(?=.*\]\]_)     // if the string contains ]]_
.+              // get the entire string
$               // end of the string

Demo here

Answer 4

I'm using this pattern just as an example.
The goal here is to use the capturing parenthesis. If the pattern matches, you will find your captured string in index n°1 in the matches array.

Example :

    $pattern = '#_(\[\[[0-9]+\]\])_#';
    $result  = preg_match_all($pattern, '_[[22555]]_ BLA BLA _[[999]]_', $matches);

    if (is_int($result) && $result > 0) {
        var_dump($matches[1]);
    }

OUTPUT

array(2) {
  [0]=>
  string(9) "[[22555]]"
  [1]=>
  string(7) "[[999]]"
}

Answer 5

Try using your pattern to capture the brackets [] and replacing the matches with what you captured, something like this:

$pattern = "/_(\[\[)|(\]\])_/";
$test =  "_[[2, verb//substantiv//adjektiv]]_";
$replace = preg_replace( $pattern ,"$1$2", $test );
echo $replace;

Dollar sign $ allows you to back reference what you captured with the parenthesis. $1 means the first capture group, in this case (\\[\\[) ,which means the first pair of brackets, $2 references the second pair of brackets. Because your pattern uses the | operator, only one of your capture groups will have a match, the other one will be empty.