Java regular expression finding number

Question

I'm trying to parse log using regex. Unfortunately I' stacked on following string. Trying to find line beginning with time

for example:

String p="11:33:00.0   2000           0.0     ....... #           0.0     ....... #           0.0     ....... #           0.0     ...";

I have following code:

public class Test {


public static void main(String[] args) {
  String p="11:33:00.0   2000           0.0     ....... #           0.0     ....... #           0.0     ....... #           0.0     ...";
  Pattern pat = Pattern.compile("^\\d\\d\\:\\d\\d*");
  Matcher m = pat.matcher(p);
  if (m.find()) {
    System.out.println(m.start());
    System.out.println(p.substring(m.start()));
  }

}
}

this code prints nothing even if I tried just '^\\d\\d'.

if I'm correct '^' stands for line beginning '\\d' for any digit

I also tried to replace '^' with '\\A' If i change pattern to

pat = Pattern.compile("\\d\\d");

it returns position at 6. Can somebody tell me why the first code is not working?:)

THX

Answer 1

You need to print the group index 0 inside the if block, so that it would print the matched characters.

String p="11:33:00.0   2000           0.0     ....... #           0.0     ....... #           0.0     ....... #           0.0     ...";
Pattern pat = Pattern.compile("^\\d\\d\\:\\d\\d*");
Matcher m = pat.matcher(p);
if (m.find()) {
    System.out.println(m.group(0));
}

Output:

11:33

Answer 2

On my computer your example Test.main worked fine, maybe it's dependant on JVM implementation. I think if you open then you also need to close regex with question mark:

"^\\d\\d?"