Replacing based on position in string

Is there a way using regex to replace characters in a string based on position?

For开发者_Go百科 instance, one of my rewrite rules for a project I’m working on is “replace o with ö if o is the next-to-last vowel and even numbered (counting left to right).”

So, for example:

• heabatoik would become heabatöik (o is the next-to-last vowel, as well as the fourth vowel)
• habatoik would not change (o is the next-to-last vowel, but is the third vowel)

Is this possible using preg_replace in PHP?

Starting with the beginning of the subject string, you want to match 2n + 1 vowels followed by an o, but only if the o is followed by exactly one more vowel:

$str = preg_replace(
  '/^((?:(?:[^aeiou]*[aeiou]){2})*)' .  # 2n vowels, n >= 0
    '([^aeiou]*[aeiou][^aeiou]*)' .     # odd-numbered vowel
    'o' .                               # even-numbered vowel is o
    '(?=[^aeiou]*[aeiou][^aeiou]*$)/',  # exactly one more vowel
  '$1$2ö',
  'heaeafesebatoik');

To do the same but for an odd-numbered o, match 2n leading vowels rather than 2n + 1:

$str = preg_replace(
  '/^((?:(?:[^aeiou]*[aeiou]){2})*)' .  # 2n vowels, n >= 0
    '([^aeiou]*)' .                     # followed by non-vowels
    'o' .                               # odd-numbered vowel is o
    '(?=[^aeiou]*[aeiou][^aeiou]*$)/',  # exactly one more vowel
  '$1$2ö',
  'habatoik');

If one doesn't match, then it performs no replacement, so it's safe to run them in sequence if that's what you're trying to do.

You can use preg_match_all to split the string into vowel/non-vowel parts and process that.

e.g. something like

preg_match_all("/(([aeiou])|([^aeiou]+)*/",
    $in,
    $out, PREG_PATTERN_ORDER);

Depending on your specific needs, you may need to modify the placement of ()*+? in the regex.

I like to expand on Schmitt. (I don't have enough points to add a comment, I'm not trying to steal his thunder). I would use the flag PREG_OFFSET_CAPTURE as it returns not only the vowels but also there locations. This is my solution:

const LETTER = 1;
const LOCATION = 2
$string = 'heabatoik'

preg_match_all('/[aeiou]/', $string, $in, $out, PREG_OFFSET_CAPTURE);

$lastElement = count($out) - 1; // -1 for last element index based 0

//if second last letter location is even
//and second last letter is beside last letter
if ($out[$lastElement - 1][LOCATION] % 2 == 0 &&
    $out[$lastElement - 1][LOCATION] + 1 == $out[$lastElement][LOCATION])
       substr_replace($string, 'ö', $out[$lastElement - 1][LOCATION]);

note:

print_r(preg_match_all('/[aeiou]/', 'heabatoik', $in, $out, PREG_OFFSET_CAPTURE));
Array
(
    [0] => Array
        (
            [0] => Array
                (
                    [0] => e
                    [1] => 1
                )

            [1] => Array
                (
                    [0] => a
                    [1] => 2
                )

            [2] => Array
                (
                    [0] => a
                    [1] => 4
                )

            [3] => Array
                (
                    [0] => o
                    [1] => 6
                )

            [4] => Array
                (
                    [0] => i
                    [1] => 7
                )
        )
)

This is how I would do it:

$str = 'heabatoik';

$vowels = preg_replace('#[^aeiou]+#i', '', $str);
$length = strlen($vowels);
if ( $length % 2 && $vowels[$length - 2] == 'o' ) {
    $str = preg_replace('#o([^o]+)$#', 'ö$1', $str);
}