...

Nope.  That will match strings that contain runs of 5 or more digits.
He wants runs of exactly 5 digits.

Here's a way to fix that so it won't incorrectly match "foo123456bar"
and similar strings:

         Pattern p = Pattern.compile(".*[^0-9]\\d{5}[^0-9].*");
         Matcher m = p.matcher("a"+str+"a");

The "a"+str+"a" ensures that we do not have to worry about the special
cases of the 5 digits being the first or last digits of the string, and
the change to the regular expression ensures that the 5 digits are a run
of exactly 5 digits, by requiring that they be bordered by a non-digit.

I suspect it's more efficient too. In general
I'd expect smaller regexps to be cheaper

  BugBear

What Hendrik said and Matcher.find() is also handy if you want to find
multiple matching subsequences.

import java.util.regex.*;

public class test2 {
    public static void main(String[] args) {
        String str = ";alsdlkajsdf 32344 lkls3 3323  llsdfsdf 12345";
        Pattern p = Pattern.compile("\\d{5}");
        Matcher m = p.matcher(str);

        while (m.find())
            System.out.println(m.group());
    }
}