java regex quantifiers

Question

I have a string like

String string = \"number0 foobar number1 foofoo number2 bar bar bar bar number3 foobar\";

I need a regex to give me th

Answer 1

Why don't you just match for number\\d+, query the match location, and do the String splitting yourself?

Answer 2

because .* is a greedy pattern. use .*? instead of .*

Pattern pattern = Pattern.compile("number\\d+(.*?)(number\\d+)");
Matcher matcher = pattern.matcher(string);
while(matcher.find();){
    out(matcher.group());
}

Answer 3

Pattern pattern = Pattern.compile("\\w+\\d(\\s\\w+)\1*");
Matcher matcher = pattern.matcher(string);

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

Answer 4

(.*) part of your regex is greedy, therefore it eats everything from that point to the end of the string. Change to non-greedy variant: (.*)?

http://docs.oracle.com/javase/tutorial/essential/regex/quant.html

Answer 5

If "foobar" is just an example and really you mean "any word" use the following pattern: (number\\d+)\s+(\\w+)

Answer 6

So you want number (+ an integer) followed by anything until the next number (or end of string), right?

Then you need to tell that to the regex engine:

Pattern pattern = Pattern.compile("number\\d+(?:(?!number).)*");

In your regex, the .* matched as much as it could - everything until the end of the string. Also, you made the second part (number\\d+)? part of the match itself.

Explanation of my solution:

number    # Match "number"
\d+       # Match one of more digits
(?:       # Match...
 (?!      #  (as long as we're not right at the start of the text
  number  #   "number"
 )        #  )
 .        # any character
)*        # Repeat as needed.