View Full Version : Java Regex Question

December 2nd, 2008, 08:30 PM
In Perl, I can do a regular expression like this:

if($_ =~ /\s[\d:\d] = \d+/)
pinname = $1;
startindex = $2;
endindex = $3;
bitsettings = $4;

So for example, the pinname corresponds to the \s string and is given index 1 ($1) because it was the first subpart of the regular expression match. The first \d of the index was matched second and was assigned to $2, and so on... Is there a way to do exactly this in Java? I have looked at Java's Regex and pattern classes, but I don't see anything that can do this. In fact, for the above example, I don't think Java's regex class will save me much work over the simpler string split method.

December 2nd, 2008, 09:46 PM
Yes, you can do it (more or less: it's done a bit differently):

import javax.regex.*;

public class Test {
public static Pattern [] my_regexps= new Pattern [] {
Pattern.compile('/^java$/'/*, optional bit mask containing flags */),
Pattern.compile('/\\w+/'/*, optional bit mask containing flags */)
public static void main (String [] args) {
Matcher m;
int i=0;
for(String s : args) {
for(Pattern p: Test.my_regexps) {
m=p.matcher(s); // evaluate s against m.
if(m.matches) {
System.out.println("The pattern \""+p.toString()+"\" matches the argument \""+s+"\"!");
System.err.println("No pattern matches the argument \""+s+"\"!");

Then there's things like getting groups/sub expressions, group/sub expression count, etc. etc all embodied within the Matcher class. However if you're looking for simple parsing (does it match or does it not/ split around a regex) you'll find that java.util.Scanner and java.lang.String got a few handy tricks built-in.