How write in java regex any string of characters?

Question

I have a text and I'd like to write a regular expression to extract the string after second @ . For example: @ some text with letter, digit 123 1234 and symbols {[ @text_to_extract . How would I write a regular expression to extract only the string after second @ . This code seems like a step in the right direction:

Pattern p = Pattern.compile("@@(.+?)");
Matcher m = p.matcher("asdasdas@@textToExtract");

This works when text between @ is empty, but how do I specify any text in a regex?

Pattern.compile("@(*)@(.+?)"); ?

Edited:
One more condition, text can be between @ and @ but doesn't have to.

Answer 1

• Don't capture the first group
• Change the plain * to .* .
• Make the second wildcard greedy, since it will otherwise capture only a single character

---

Pattern.compile("@.*@(.+)");

Answer 2

The "non-greedy" operator should be removed. (.*?) should be (.*) ... Otherwise you match just the minimum of the text after the second @. Definitely need a "." in front of the *. It means "0 or more of the proceeding character. Actually, maybe you want [^@]* instead... so it matches anything but the at symbol.. so you're guaranteed to get everything, even if . doesn't match newlines. Anyway, here's working code.

import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.regex.*;

class Ideone {
    public static void main(String[] args) throws java.lang.Exception {
        // Pattern p = Pattern.compile("@(*)@(.+?)");
        Pattern p = Pattern.compile("@.*@(.+)");
        Matcher m = p.matcher("asdasdas@@textToExtract");

        while (m.find()) {
            System.out.println(m.group(1));
        }
    }
}

Play with the code here: http://ideone.com/rxB5Zy

Answer 3

你应该这样

Matcher m =Pattern.compile("^[^@]*@[^@]*@([^@]*)").matcher(input);