Question

Here I am using regex to catch first 15 characters to match, and while using substring I have to use (0,matcher.start()) only wherein I should get 15 only, kindly help me out.

   String test = "hello world this is example";
        Pattern p = Pattern.compile(".{15}");

    //can't change the code below
    //can only make changes to pattern 


        Matcher m=p.matches(test);
        matcher.find(){
            String string = test.substring(0, m.start());
        }

    //here it is escaping the first 15 characters but I need them
    //the m.start() here is giving 0 but it should give 15

Answer 1

If possible, I agree with @jlordo’s comment: Use String string = test.substring(0, 15);

If you are forced to pass through the bottom code snippet which you marked as unchangeable, there is a workaround.
(Well that depends… If you are stuck with an unchangeable code snippet does not even compile… You’re gonna have a bad time)

If you really need a regex that always returns 15 for m.start() you can use the regex lookbehind concept.

        String test = "hello world this is example";
        Pattern p = Pattern.compile("(?<=.{15}).");

        Matcher m=p.matcher(test);
        m.find();
        System.out.println(m.start());

Granted that the test input string is at least 16 characters long, this will always return 15 for m.start().
The regex is supposed to be read as “Any character (the last .) preceded by (the (?<=) lookbehind operator) exactly 15 any characters (the .{15})”.

(?<=foo) is a lookbehind operator specifying that whatever regex comes after must be preceded by “foo”.
E.g. the regex: (?<=foo)bar
Will match the bar in: foobar
But not the bar in: wunderbar