Java - Regex Match Multiple Words

You could try this:

^when is (.*)$

This will find a string that starts with when is and capture everything else to the end of the line.

The regex will return one group. You can access it like so:

String line = "when is Veteran's Day.";
Pattern pattern = Pattern.compile("^when is (.*)$");
Matcher matcher = pattern.matcher(line);
while (matcher.find()) {
    System.out.println("group 1: " + matcher.group(1));
    System.out.println("group 2: " + matcher.group(2));
}

And the output should be:

group 1: when is Veteran's Day.
group 2: Veteran's Day.

If you want to allow whitespace to be matched, you should explicitly allow whitespace.

([\w\s]+)

However, roydukkey's solution will work if you want to capture everything after when is.

Don't use regular expressions when you don't need to!! Although the theory of regular expressions is beautiful in the thought that you can have a string do code operations for you, it is very memory inefficient for simple use cases.

If you are trying to get the word after "when is" ending by a space, you could do something like this:

String start = "when is ";
String end = " ";
int startLocation = fullString.indexOf(start) + start.length();
String afterStart = fullString.substring(startLocation, fullString.length());
String word = afterStart.substring(0, afterStart.indexOf(end));

If you know the last word is Day, you can just make end = "Day" and add the length of that string of where to end the second substring.

You can express this as a character class and include spaces in it: when is ([\w ]+).

\w only includes word characters, which doesn't include spaces. Use [\w ]+ instead.