Java regular expression find substring

Question

I'm trying to find a specific word into a string in Java. I developed a function aims to return the found string. THis is what I wrotte for now:

public static String getValueByregexExpr (String str, String regexExpr) {
    Pattern regex = Pattern.compile (regexExpr, Pattern.DOTALL);
    Matcher matcher1 = regex.matcher (str);
    if (matcher1.find ()) {
        if (matcher1.groupCount () != 0 && matcher1.group (1) != null) {
            for (int i = 0; i <= matcher1.groupCount (); i++) {
                System.out.println ("matcher " + i + " for regex " + regexExpr + "= " + matcher1.group (i));
            }
            return matcher1.group (1);
        }
        return regexExpr;
    }
    return null;
}

My issue is the following, I wish to find a regex able to fill in the group(1) with the word I'm looking for. But for now this code:

public static void main (String[] args) {

    String str = "HELLO_WORLD_123456 TEst";

    System.out.println ("First test");
    String regex1 = ".*WORLD.*";
    String matchedString = Util.getValueByregexExpr (str, regex1);
    //Here, I want to obtain matchedString = WORLD
    if (matchedString == null) {
        System.out.println ("matchedString null");
    } else if (matchedString.equals (regex1)) {
        System.out.println ("String found but empty group(1)");
    } else {
        System.out.println ("Result : " + matchedString);
    }

    //Here, I want to obtain matchedString = WORLD_123456
    System.out.println ("\nSecond test");
    String regex2 = "WORLD_([^_]+)";
    matchedString = Util.getValueByregexExpr (str, regex2);
    if (matchedString == null) {
        System.out.println ("regex " + regex2 + " matchedString null");
    } else if (matchedString == regex2) {
        System.out.println ("regex " + regex2 + " String found but empty group(1)");
    } else {
        System.out.println ("regex " + regex2 + " Result : " + matchedString);
    }

}

Give me the output:

First test:
regex .*WORLD.* String found but empty group(1)

Second test:
matcher 0 for regex WORLD_([^_]+)= WORLD_123456
matcher 1 for regex WORLD_([^_]+)= 123456
regex WORLD_([^_]+) Result : 123456

First, is there any regular expression whitch can returns: - WORLD for the first test - WORLD_123456 for the second test

Second, I thought at the begining that every result will be set into group(1) as long as you have only one result. But I'm obviously wrong given test 2 result. Could someone give me more information about it?

Thank you for your help.

@ShrinivasShukla: The second regex "(WORLD_[^_]+)" will return WORLD_123456 TEst. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Aug 7, 2015 at 8:48
Do this. String regex1 = ".*(WORLD).*"; and String regex2 = "(WORLD_[^_\\s]+)"; — Shrinivas Shukla
– Shrinivas Shukla, Commented Aug 7, 2015 at 8:51

Wiktor Stribiżew · Accepted Answer · 2015-08-07 08:53:38Z

1

To fix the first one, just add the capturing group:

String regex1 = ".*(WORLD).*";

To fix the second, add whitespace to the character class:

String regex2 = "(WORLD_[^_\\s]+)";

See demo

The main reason your first part of code was not working as expected is the missing capturing groups that your getValueByregexExpr is checking for. The second one returned the part of stirng captured with the ([^_]+) regex part.

edited Aug 7, 2015 at 8:53

answered Aug 7, 2015 at 8:48

Wiktor Stribiżew

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1 Comment

loic Over a year ago

I didn't realize the importance of the "()". Using your regex it is working now, and all the information I need are set into group(1). Thank you.

Shrinivas Shukla · Accepted Answer · 2015-08-07 08:49:04Z

0

In regex, everything inside () becomes a group.

Correct you regex.

String regex1 = ".*(WORLD).*";

String regex2 = "(WORLD_[^_\\s]+)";

answered Aug 7, 2015 at 8:49

Shrinivas Shukla

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