Java Regex capture nested matches

Question

I am having trouble with regex here.

Say i have this input:

608094.21.1.2014.TELE.&BIG00Z.1.1.GBP

My regex looks like this

(\d\d\d\d\.\d?\d\.\d?\d)|(\d?\d\.\d?\d\.\d?\d?\d\d)

I want to extract the date 21.1.2014 out of the string, but all i get is

8094.21.1

I think my problem here is, that 21.1.2014 starts within the (wrong) match before. Is there a simple way to make the matcher look for the next match not after the end of the match before but one character after the beginning of the match before?

What you want is to do something like "\d*(\d{4}.\d{1,2}.\d{1,2})|(\d{1,2}\.\d{1,2}.\d{2,4})" — Secespitus
– Secespitus, Commented Jul 7, 2017 at 10:47

Federico Piazza · Accepted Answer · 2017-07-07 16:11:41Z

1

You could use a regex like this:

\d{1,2}\.\d{1,2}\.\d{4}

Working demo

Or shorten it and use:

(\d{1,2}\.){2}\d{4}

answered Jul 7, 2017 at 16:11

Federico Piazza

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Toto · Accepted Answer · 2017-07-07 10:47:06Z

1

If the date is always surrounded by dot:

\.(\d\d\d\d\.\d?\d\.\d?\d|\d?\d\.\d?\d\.\d?\d?\d\d)\.

answered Jul 7, 2017 at 10:47

Toto

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piet.t · Accepted Answer · 2017-07-12 09:22:07Z

0

I hope this will help you.

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Test {
    public static void main(String[] args) {
        String x = "608094.21.1.2014.TELE.&BIG00Z.1.1.GBP";
        String pattern = "[0-9]{2}.[0-9]{1}.[0-9]{4}";

        // Create a Pattern object
        Pattern r = Pattern.compile(pattern);

        // Now create matcher object.
        Matcher m = r.matcher(x);
        if (m.find( )) {
            System.out.println("Found value: " + m.group() );
        }else {
            System.out.println("NO MATCH");
        }
    }

edited Jul 12, 2017 at 9:22

piet.t

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answered Jul 7, 2017 at 10:54

Sudha Velan

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