Pattern/Matcher group() to obtain substring in Java?

wufoo picture wufoo · Dec 7, 2012 · Viewed 31.7k times · Source

UPDATE: Thanks for all the great responses! I tried many different regex patterns but didn't understand why m.matches() was not doing what I think it should be doing. When I switched to m.find() instead, as well as adjusting the regex pattern, I was able to get somewhere.


I'd like to match a pattern in a Java string and then extract the portion matched using a regex (like Perl's $& operator).

This is my source string "s": DTSTART;TZID=America/Mexico_City:20121125T153000 I want to extract the portion "America/Mexico_City".

I thought I could use Pattern and Matcher and then extract using m.group() but it's not working as I expected. I've tried monkeying with different regex strings and the only thing that seems to hit on m.matches() is ".*TZID.*" which is pointless as it just returns the whole string. Could someone enlighten me?

 Pattern p = Pattern.compile ("TZID*:"); // <- change to "TZID=([^:]*):"
 Matcher m = p.matcher (s);
 if (m.matches ()) // <- change to m.find()
    Log.d (TAG, "looking at " + m.group ()); // <- change to m.group(1)

Answer

hoaz picture hoaz · Dec 7, 2012

You use m.match() that tries to match the whole string, if you will use m.find(), it will search for the match inside, also I improved a bit your regexp to exclude TZID prefix using zero-width look behind:

     Pattern p = Pattern.compile("(?<=TZID=)[^:]+"); //
     Matcher m = p.matcher ("DTSTART;TZID=America/Mexico_City:20121125T153000");
     if (m.find()) {
         System.out.println(m.group());
     }