正则表达式：重复字符的最后一次出现

Question

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

eg 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: :-)

Answer 1

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 .

Answer 2

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

Segment	Description
[^aeiou]+	one or more of anything BUT vowels
[aeiou]+	one or more vowels
(?=[^aeiou])	will be a match if it is followed by anything BUT a vowel
|	OR
.*[aeiou](?=\b)	zero or more of any character followed by a vowel and it needs to be followed by a non-word

 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));