Regular expression to allow decimal places including .123 and 123. etc in Java

Question

Friends

I have the following Regex "^[0-9]{1,12}+(\\\\.[0-9]{1,4})?$" that allows the following values

123456789012.1234 which is a valid decimal value {12,4}

123456789012 which is a valid integer value etc

but it doesn't allow value like .1235 etc , how should I modify above Regex so that it also allows values like .123 and 123. etc

Answer 1

I guess,

^(?:[0-9]{1,12}(?:\\.[0-9]{0,4})?|\\.[0-9]{0,4})$

might be somewhat close.

TEST

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


public class RegularExpression{

    public static void main(String[] args){

                
        final String regex = "^(?:[0-9]{1,12}(?:\\.[0-9]{0,4})?|\\.[0-9]{0,4})$";
        final String string = "123456789012.1234\n"
             + "123456789012\n"
             + ".1235\n"
             + ".123\n"
             + "123.\n"
             + ".12345\n"
             + "0.12345";

        final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
        final Matcher matcher = pattern.matcher(string);

        while (matcher.find()) {
            System.out.println("Full match: " + matcher.group(0));
            for (int i = 1; i <= matcher.groupCount(); i++) {
                System.out.println("Group " + i + ": " + matcher.group(i));
            }
        }


    }
}

OUTPUT

Full match: 123456789012.1234
Full match: 123456789012
Full match: .1235
Full match: .123
Full match: 123.

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If you wish to simplify/update/explore the expression, it's been explained on the top right panel of regex101.com . You can watch the matching steps or modify them in this debugger link , if you'd be interested. The debugger demonstrates that how a RegEx engine might step by step consume some sample input strings and would perform the matching process.

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RegEx Circuit

jex.im visualizes regular expressions: