To match a bar code: Add regex to a given string, whithout changing the string itself

Question

I get a string in this format: 00123456

The string has always 8 digits, starts always with 2 zeros, and the other 6 digits can be anything from 1-9.

I have to compare this string to another string, which has the same structure, but without the 2 zeros. So, 00123456 should be equal to 123456. The problem is, that I don't have the possibility to change the strings (eg delete the zeros). The only thing I could do, is to add regex before or after the first string.

So in Java code, I need something like this, whithout changing the first string itself. I can only add something before and/or after the string:

    String givenStringWithRegex = "(ignore zeros)00123456";
    String userInput = "123456";
    if (userInput.matches(givenStringWithRegex)) {
        // Everything is fine :)
    }

Thank you in advance :)

Edit: Thank you for your answers so far, but the problem is, that I cannot change the code of the client application. The client app does userInputString.matches(givenStringWithRegex) and I cannot change this. Thats why I need another solution, for example adding regex to the given string without changing the string itself.

Edit2: This is the code in the client application and I can nothing change here. The "regex" variable is the given string and "scannedValue" is the user input (123456). The given string is in my concrete example 00123456. Like I wrote before, I cannot manipulate 00123456. I can only add regex before or after it, for example ???00123456???. And 00???123456 is not possible.

I can add this regex in a configuration file of the superior system, so no code change would be needed. Thats why I asked for a regex solution, because I don't want to change code in superior system and client application.

public final class RegexValidator implements StringScanValidator {

    private final String regex;

    public RegexValidator(final String regex) {
        this.regex = regex;
    }

    @Override
    public StringScanValidatorResult validate(final String scannedValue) {
        return scannedValue.matches(regex) ? new StringScanValidatorResult(true, scannedValue)
            : new StringScanValidatorResult(false, scannedValue);
    }
}

Answer 1

Just use substring() like this :

public static void main(String[] args) {
    String givenStringWithRegex = "00123456"; // always 8 digits
    String userInput = "123456"; //so, always 6 digits
    System.out.println(givenStringWithRegex.substring(2, 8).equals( // do substring for digits 2 to 8
            userInput));
}

O/P :

true

Answer 2

String givenString = "00123456".substring(2); 
String userInput = "123456";

 if(userInput.equals(givenString) { // Everything is fine :) }

Substringing will fit your need. It will cut a portion of the string that is relevant to you.

Answer 3

Can you do something like this:

String givenString = "00123456";
String userInput = "123456";
if (("00"+userInput).equals(givenString)) {
       // Everything is fine :)
}

Answer 4

Apart from your problem you never can change String s in java: they are immutable. But you can make any number of copies of an original String. So you can use substring to cut of the 0 s without changing the original String:

String original = "00123456";
String compare  = "123456";
String copy = original.substring(2);
if(copy.equals(compare)) {...}

An other way would be to prepend user-input with zeros:

String original = "00123456";
String compare  = "123456";
if(original.equals("00"+compare)) {...}

So using regular expresion seems like terrible overkill!

Answer 5

Strings are immutable. You could do something like:

String givenStringWithRegex = "(ignore zeros)00123456";
String userInput = "123456";
if (("00"+userInput).equals(givenStringWithRegex)) {
    //its equal and none of your strings are modified as string are immutable
}

Answer 6

public static String regexStr(String str) {
    Pattern p =
        Pattern.compile("00(\\d{6})", Pattern.CASE_INSENSITIVE | Pattern.DOTALL); //start with 00 then followed by 6 digits
    Matcher m = p.matcher(str);
    if (m.find()) {
        String found= m.group(1);
        return found;
    }
    return null;
}

public static void main(String[] args) throws ParseException {
    String givenStringWithRegex = "00123456";
    String userInput = "123456";
    if (userInput.equalsIgnoreCase(regexStr(givenStringWithRegex))) {
        // Everything is fine :)
        System.out.println("Everything is fine");
    }
}

Answer 7

This is the code in the client application and I can nothing change here. The "regex" variable is the given string and "scannedValue" is the user input (123456). The given string is in my concrete example 00123456. Like I wrote before, I cannot manipulate 00123456. I can only add regex before or after it, for example ???00123456???. And 00???123456 is not possible.

 scannedValue.matches(regex) 

The regex rules make the fulfillment of your wish impossible.