CODEWITHSUNDEEP
javaregex
sala
sala

Reputation: 181

Java regular expression finding number

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

Upvotes: 0

Views: 63

Answers (2)

Kristjan Veskimäe
Kristjan Veskimäe

Reputation: 992

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?"

Upvotes: 0

Avinash Raj
Avinash Raj

Reputation: 174696

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

Upvotes: 1

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