Find the last match with Java regex matcher

Question

I'm trying to get the last result of a match without having to cycle through .find()

Here's my code:

String in = "num 123 num 1 num 698 num 19238 num 2134"; Pattern p = Pattern.compile("num '([0-9]+) "); Matcher m = p.matcher(in);  if (m.find()) {      in = m.group(1); } 

That will give me the first result. How do I find the LAST match without cycling through a potentionally huge list?

Answer 1

You could prepend .* to your regex, which will greedily consume all characters up to the last match:

import java.util.regex.*;  class Test {   public static void main (String[] args) {     String in = "num 123 num 1 num 698 num 19238 num 2134";     Pattern p = Pattern.compile(".*num ([0-9]+)");     Matcher m = p.matcher(in);     if(m.find()) {       System.out.println(m.group(1));     }   } } 

Prints:

2134 

You could also reverse the string as well as change your regex to match the reverse instead:

import java.util.regex.*;  class Test {   public static void main (String[] args) {     String in = "num 123 num 1 num 698 num 19238 num 2134";     Pattern p = Pattern.compile("([0-9]+) mun");     Matcher m = p.matcher(new StringBuilder(in).reverse());     if(m.find()) {       System.out.println(new StringBuilder(m.group(1)).reverse());     }   } } 

But neither solution is better than just looping through all matches using while (m.find()), IMO.

Answer 2

To get the last match even this works and not sure why this was not mentioned earlier:

String in = "num 123 num 1 num 698 num 19238 num 2134"; Pattern p = Pattern.compile("num '([0-9]+) "); Matcher m = p.matcher(in); if (m.find()) {   in= m.group(m.groupCount()); }