Is there a way using regex to replace characters in a string based on position?
For 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.
+ at that last [^aeiou]+ and not a *?
heaeafesebatoik gives me fesebatöikYou 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);
}
