Regex for file extension

[gapper]

Hi guys,

I have file paths like:

C:\dir\filename.ext

I want to get the extension part form it. I tried the following but failed:

.[A-Za-z]*

Please advice. I would appreciate an early response to this.

[tim]

Hello

Sorry for the delay. I had to study this topic before I could help you. Using Regex is a bit more complicated than your proposal. I created a program that uses the Regex package to do as you asked.

package p1;

import java.util.regex.*;

public class Main {
    public static void main(String[] args) {
        /*
        Regex: (?<=\\.).*$
        1.)     (?<=\\.)    means       lookbehind excluding "." and add
        2.)     .*          means       every character until
        3.)     $           meabs       end of string
        */
        
        String
                regex = "(?<=\\.).*$",
                input = "C:\\dir\\filename.ext";
        Pattern pattern = Pattern.compile(regex);
        Matcher matcher = pattern.matcher(input);
        String output = "no match";
        if (matcher.find()){
            output = matcher.group();
        }
        System.out.println("\t Output: " + output);
    }
}

This will output

         Output: ext

For help on Regex syntax see this link.

This stuff is excellent! I hope that helped you. :D