Rular expression matching with java

linuxmandrake · 04-13-2006, 07:19 AM

String x = s.replaceAll("[a-zA-Z]",s);
public static void main(String[] args) {
Highscores high = new Highscores();

System.out.println(high.process("Steven 500"));

}
gives me Steven 500Steven 500Steven 500Steven 500Steven 500Steven 500 500
why? According to http://java.sun.com/developer/techni...ases/1.4regex/ my regular expression is valid

xhi · 04-13-2006, 11:35 AM

well since you omitted some code i am not sure what HighScores.process() does.. however i would guess that you are replacing every letter (you have 6 of them) with the string 'Steven 500' (which is printed 6 times)..

taylor_venable · 04-13-2006, 11:35 AM

It looks like you want to do a regex-powered replacement on some string, right? This code using java.util.regex works for me:

Code:

import java.util.regex.*;

public class StringMatch {
    public static void main(String[] args) {
        Pattern p = Pattern.compile("foo");
        for (int i = 0; i < args.length; i += 1) {
            System.out.println(p.matcher(args[i]).replaceAll("bar"));
        }
    }
}

It goes through each argument and replaces each occurrence of "foo" with "bar".

First you have to create a regex Pattern object using the Pattern class' static compile() method. Then call p.matcher(String) to get a Matcher object from that pattern, ready to process whatever string you pass as an argument. Then you can call a function on the Matcher, which does whatever you want. replaceAll(String) does a substitution and returns a new string. matches() just returns true or false. You can even get grouped-match results using the group(int) function.