Problem with formatting a regular expression

Question

I am trying to format this regular expression into a String pattern

    (^(234\d{7,12})$)|(\b(234\d{7,12}\b\s*,\s*)(\b(234\d{7,12})\b)*)

This is an accurate regex (a has been validated in regexpal.com as being so)

But when I try it in java, it shows errors. And even if I escape it using //, It still dosen't give an accurate Logic. Please. How can I solve this.

package MCast;

import java.util.StringTokenizer;
import java.util.regex.*;
import javax.swing.JOptionPane;

/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author nnanna
 */


public class Verify {

    private static final String STOP = "STOP";
    private static final String VALID = "Valid Java Identifier";
    private static final String INVALID = " Not Valid Number Format: Must be of the form 23400000023";
    private static final String VALID_IDENTIFIER_PATTERN = "(^(234\\d{7,12})$)|(\\b(234\\d{7,12}\\b\\s*,\\s*)(\\b(234\\d{7,12})\\b)*)";
//    private static final String VALID_IDENTIFIER_PATTERN2 = "[[2-3][2-3][3-4][0-9]*[ ][2-3][2-3][3-4][0-9]*]*";//[,][2[0-9]]{11}]*";
    static String str;
    boolean reply;

    public Verify() {
    }

    public int countNo(String stringToCount) {
        int j = stringToCount.length();

        int count = 0;

        for (int i = 0; i < j; i++) {
            if (stringToCount.charAt(i) == ',') {
                count += 1;
            }
        }
//        System.out.println(count);
        return count + 1;
    }

    public boolean  pattern(String str){

         Matcher match;
    Pattern pattern = Pattern.compile(VALID_IDENTIFIER_PATTERN);

            match = pattern.matcher(str);
            if (match.matches()) {
                reply = true;
                JOptionPane.showMessageDialog(null, str + ":\n" + reply + "\n" +countNo(str));
            } else {
                reply = false;
                JOptionPane.showMessageDialog(null, str + ":\n" + reply + "\n");

            }
            return reply;

    }

    public static void main(String args[]){
         Verify a = new Verify();
         String test1 = "23439869450";
         String test2 = "23439869450,23439869450";
         String test3 = "23439869450,23439869450,23439869450";
         String test4 = "23439869450,23439869450,23439869450,23439869450,23439869450,23439869450";
         String test5 = "07039869450,23439869450,23439869450,23439869450,23439869450,23439869450";
//         a.pattern(test1);
//         System.out.println(a.countNo(test1));

         a.pattern(test3);
         System.out.println(a.countNo(test2));
         System.out.println(a.pattern(test1));
         System.out.println(a.pattern(test2));
         System.out.println(a.pattern(test3));
         System.out.println(a.pattern(test4));
         System.out.println(a.pattern(test4));
//
//         a.pattern(null);
//         System.out.println(a.countNo(test1));
    }
}

@dacwe, I tried that, It didnt work. The escape sequence out to be \\ — nnanna
– nnanna, Commented Sep 16, 2011 at 11:05
But even when I use \\ It doesn't carry out the correct logic — nnanna
– nnanna, Commented Sep 16, 2011 at 11:06
I really dont understand. I should probably attache the code snippet for you guys to take a look — nnanna
– nnanna, Commented Sep 16, 2011 at 11:12

Tim Pietzcker · Accepted Answer · 2011-09-16 12:16:45Z

3

You need to double the backslashes. And your regex isn't doing what you think it is. Use this:

Pattern regex = Pattern.compile(
    "^\n" +
    "234\\d{7,12}\\s*,\\s*234\\d{7,12}               # match a pair\n" +
    "(?:\\s*,\\s*234\\d{7,12}\\s*,\\s*234\\d{7,12})* # optionally match more pairs\n" +
    "$", 
    Pattern.COMMENTS);
Matcher regexMatcher = regex.matcher(subjectString);
foundMatch = regexMatcher.matches();

This allows pairs of numbers starting with 234; 10-15 digits long. All numbers must be comma-separated.

edited Sep 16, 2011 at 12:16

answered Sep 16, 2011 at 10:54

Tim Pietzcker

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5 Comments

nnanna Over a year ago

You can test it ou in www.regex.com with the following numbers to see what i mean

nnanna Over a year ago

23439869450 23439869450,23439869450 23439869450,23439869450,23439869450 23439869450,23439869450,23439869450,23439869450,23439869450,23439869450 07039869450,23439869450,23439869450,23439869450,23439869450,23439869450

nnanna Over a year ago

It is supposed to accept any number that starts with (1) 234 (2) Is in the range of 10-15 in length (3) Allow commas(,) and a comma with space(, ) after the first for number for subsequent numbers

Tim Pietzcker Over a year ago

Your regex is written to mean "match a string that consists only of one number starting with 234 etc., or match pairs of such numbers if they are separated by commas".

nnanna Over a year ago

Match pairs of numbers starting with 234... and allow to occur 1 or more times

epoch · Accepted Answer · 2011-09-16 10:59:13Z

1

don't you mean escaping the \ with \\ and not // ex : \\d

edited Sep 16, 2011 at 10:59

answered Sep 16, 2011 at 10:53

epoch

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2 Comments

nnanna Over a year ago

ya, I mean't that. But does not seem to give an accurate logic

nnanna Over a year ago

Just pasted the code. In the main method, tests 1-test 4 is supposed to show true while test 5 should show false since one of the numbers doesn't begin with 234

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