So, I am looking for a Regex that is able to match with every maximal non-empty substring of consonants followed by a maximal non-empty substring of vowels in a String

e.g. In the following strings, you can see all expected matches:

"zcdbadaerfe" = {"zcdba", "dae", "rfe"}

"foubsyudba" = {"fou", "bsyu", "dba"}

I am very close! This is the regex I have managed to come up with so far:

([^aeiou].*?[aeiou]) 

It returns the expected matches except for it only returns the first of any repeating lengths of vowels, for example:

String: "cccaaabbee"

Expected Matches: {"cccaaa", "bbee"}

Actual Matches: {"ccca", "bbe"}

I want to figure out how I can include the last found vowel character that comes before (a) a constant or (b) the end of the string.

Thanks! :-)

CodePudding user response：

Your pattern is slightly off. I suggest using this version:

[b-df-hj-np-tv-z] [aeiou] 

This pattern says to match:

• [b-df-hj-np-tv-z] a lowercase non vowel, one or more times
• [aeiou] followed by a lowercase vowel, one or more times

Here is a working demo.

CodePudding user response：

const rgx = /[^aeiou] [aeiou] (?=[^aeiou])|.*[aeiou](?=\b)/g;

function lastVowel(str) {
  const rgx = /[^aeiou] [aeiou] (?=[^aeiou])|.*[aeiou](?=\b)/g;
  return [...str.matchAll(rgx)].flat();
}

const str1 = "cccaaabbee";
const str2 = "zcdbadaerfe";
const str3 = "foubsyudba";

console.log(lastVowel(str1));
console.log(lastVowel(str2));
console.log(lastVowel(str3));