simple java regex throwing illegalstateexception
illegalstateexception, java, regex
Solution
You need to invoke `m.find()` or `m.matches()` first to be able to use `m.group`.
- `find` can be used to find each substring that matches your pattern (used mainly in situations where there is more than one match)
- `matches` will check if entire string matches your pattern so you wont even need to add `^` and `$` in your pattern.
We can also use `m.lookingAt()` but for now lets skip its description (you can read it in documentation).
Problem
Im trying to do a quick sanity check... and its failing. Here is my code - ``` import java.util.regex.*; public class Tester { public static void main(String[] args) { String s = "a"; Pattern p = Pattern.compile("^(a)$"); Matcher m = p.matcher(s); System.out.println("group 1: " +m.group(1)); } } ``` And what I would expect is to see `group 1: a`. But instead I get an `IllegalStateException: no match found` and I have no idea why. Edit: I also tries printing out `groupCount()` and it says there is 1.