Java pattern matching - perl like

I have an XML file and I try to match something like this:
Pattern p = Pattern.compile("category code=\\S{1,} validFrom=\\S{1,}");
Matcher m = p.matcher (content);
while (m.find ()) {
System.out.println ("Found " + m.group ());
This code works and extracts groups according to the pattern but I want to extract only the value.
Exemple:
the string is: category code=23 validFrom=2011-01-01
I want to extract only *23* and *2011-01-01*.
Can I do this without manualy parsing the string?
Thank you.

XPath is the best solution for this. If you must have regex, see java.until.regex.

Similar Messages

  • Java Pattern Matcher (Pattern.class bug? Stuck in Infinite Loop)

    Hi,
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    java.util.regex.Matcher[pattern=[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*@(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+(?:[A-Z]{2}|com|org|net|biz|info) region=0,353648 lastmatch=
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    atom.study(info);
    info.minLength = minL;
    info.deterministic = false;
    return next.study(info);
    } else {
    atom.study(info);
    return next.study(info);
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    // the loop count should NOT be incremented
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    matcher.locals[countIndex] = count;
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    if (next.match(matcher, i, seq))
    return true;
    if (count < cmax) {
    matcher.locals[countIndex] = count + 1;
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    The Java Pattern Matcher is getting stuck in the following code...
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  • Java pattern matching

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    The ?? tell the regex to choose the minimum needed to make the group happy. So the \b?? wont take punctuation unless no other groups want it.Ummmm... no.

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    If you use just / it misinterprets it and it ruins
    your " " tags for a string. I don't think so. '/' is not a special character for Java regex, nor for Java String.
    The reason i used
    literal is to try to force it to directly match,
    originally i thought that was the reason it wasn't
    working.That will be no problem because it enforces '.' to be treated as a dot, not as a regex 'any character'.
    Message was edited by:
    hiwa

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