A Java regular expression about finding digit string

Question

I was learning a Java regular expression tutorial online and got confused about one small program.

  // String to be scanned to find the pattern.
  String line = "This order was places for QT3000! OK?";
  String pattern = "(.*)(\\d+)(.*)";

  // Create a Pattern object
  Pattern r = Pattern.compile(pattern);

  // Now create matcher object.
  Matcher m = r.matcher(line);
  if (m.find( )) {
     System.out.println("Found value: " + m.group(0) );
     System.out.println("Found value: " + m.group(1) );
     System.out.println("Found value: " + m.group(2) );
  }

And the results printed out are:

Found value: This order was places for QT3000! OK?

Found value: This order was places for QT300

Found value: 0

I have no idea why the group(1) gets value the above value? Why it stops before the last zero of 'QT3000'?

Thank you very much!

the zero you're missing in group 1 is right there in group 2... group 3 will be ! OK?. — cha0site
– cha0site, Commented Jul 11, 2012 at 14:33

David M · Accepted Answer · 2012-07-11 14:32:35Z

2

The first group of (.*) (this is index 1 - index 0 is the overall regular expression) is a greedy match. It captures as much as it can while letting the overall expression still match. Thus it can take up to the second 0 in the string, leaving just 0 to match (\\d+). If you want different behaviour, then you should read up on greedy and non-greedy matches, or find a more appropriate pattern.

answered Jul 11, 2012 at 14:32

David M

73.2k13 gold badges164 silver badges187 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

ilomambo · Accepted Answer · 2012-07-11 14:38:49Z

0

Actually you got the group numbers wrong.

Group 0 will always be the whole string to match

Group 1 will be the match for (.*) which is called "greedy" because it will match as many characters as possible (in your case "This order was places for QT300")

Group 2 is the match for (\d+) which is the minimum possible to match the regex (in your case it is "0")

Group 3 (which you did not print) is the last (.*) and should match "! OK" ( The "?" is a special regex character, if you want to match it litterally prefix it with \)

If you want to match the 3000 on group 2 use this regex:

String pattern = "(.*?)(\\d+)(.*)";

answered Jul 11, 2012 at 14:38

ilomambo

8,36013 gold badges62 silver badges109 bronze badges

1 Comment

D.Q. Over a year ago

Yes, the pattern you provided for matching '3000' worked. And now I understand the use of '?' as well. Thanks!

Collectives™ on Stack Overflow

A Java regular expression about finding digit string

2 Answers 2

Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related