phone Number validation in java

Question

I want to validate a phone number in such Way :-

The field should allow the user to enter characters and should auto-correct. So an entry of "+1-908-528-5656" would not create an error for the user, it would just change to "19085285656".

I also want to number range between 9 to 11.

I also tried with the below code but not concluded to the final solution:

 final String PHONE_REGEX = "^\\+([0-9\\-]?){9,11}[0-9]$";
 final Pattern pattern = Pattern.compile(PHONE_REGEX);
 String phone = "+1-908-528-5656";      
 phone=phone.replaceAll("[\\-\\+]", "");
 System.out.println(phone);
 final Matcher matcher = pattern.matcher(phone);
 System.out.println(matcher.matches());

what's wrong the code you have now? Can't you just replace {9,10} with {9,11} to fix your problem? — k_g
– k_g, Commented Dec 24, 2014 at 4:59
its works but code not optimized and there should be a best(it can be done in one line) way to do it. — Devendra
– Devendra, Commented Dec 24, 2014 at 5:04

Braj · Accepted Answer · 2014-12-24 05:15:56Z

6

You can use simple String.matches(regex) to test any string against a regex pattern instead of using Pattern and Matcher classes.

Sample:

boolean isValid = phoneString.matches(regexPattern);

Find more examples

Here is the regex pattern as per your input string:

\+\d(-\d{3}){2}-\d{4}

Online demo

Better use Spring validation annotation for validation.

Example

edited Dec 24, 2014 at 5:15

answered Dec 24, 2014 at 5:08

Braj

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2 Comments

Devendra Over a year ago

Could you tell me what is difference between phoneString.matches(regexPattern); and Matcher matcher = pattern.matcher(phone); matcher.matches()

Braj Over a year ago

String.matches(regex) are used to match whole string whereas pattern and matcher can also be used for matching sub-string as well. to match whole string using pattern and matcher you have to use ^ and $ to match start and end of the string respectively.

Jaspreet Kaur · Accepted Answer · 2017-01-21 09:18:22Z

4

// The Regex not validate mobile number, which is in internation format.
// The Following code work for me. 
// I have use libphonenumber library to validate Number from below link.
// http://repo1.maven.org/maven2/com/googlecode/libphonenumber/libphonenumber/8.0.1/
//  https://github.com/googlei18n/libphonenumber
// Here, is my source code.

 public boolean isMobileNumberValid(String phoneNumber)
    {
        boolean isValid = false;

        // Use the libphonenumber library to validate Number
        PhoneNumberUtil phoneUtil = PhoneNumberUtil.getInstance();
        Phonenumber.PhoneNumber swissNumberProto =null ;
        try {
            swissNumberProto = phoneUtil.parse(phoneNumber, "CH");
        } catch (NumberParseException e) {
            System.err.println("NumberParseException was thrown: " + e.toString());
        }

        if(phoneUtil.isValidNumber(swissNumberProto))
        {
            isValid = true;
        }

        // The Library failed to validate number if it contains - sign
        // thus use regex to validate Mobile Number.
        String regex = "[0-9*#+() -]*";
        Pattern pattern = Pattern.compile(regex);
        Matcher matcher = pattern.matcher(phoneNumber);

        if (matcher.matches()) {
            isValid = true;
        }
        return isValid;
    }

answered Jan 21, 2017 at 9:18

Jaspreet Kaur

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1 Comment

Luke James Over a year ago

THis should be the top answer. Using libphonenumber library is way better than regex. Especially if you need to validate international numbers. javalibs.com/artifact/com.googlecode.libphonenumber/….

Md. kamrul Hasan · Accepted Answer · 2014-12-24 06:24:33Z

1

Assuming your input field take any kind of character and you just want the digits.

 String phone = "+1-908-528-5656";
 phone=phone.replaceAll("[\\D]","");
 if(phone.length()>=9 || phone.length()<=11)
   System.out.println(phone);

answered Dec 24, 2014 at 6:24

Md. kamrul Hasan

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Comments

rohan · Accepted Answer · 2016-11-15 05:28:35Z

1

We can use String.matches(String regex)1 to validate phone numbers using java.

Sample code snippet

package regex;

public class Phone {
    private static boolean isValid(String s) {
        String regex = "\\d{3}-\\d{3}-\\d{4}"; // XXX-XXX-XXXX
        return s.matches(regex);
    }

    public static void main(String[] args) {
        System.out.println(isValid("123-456-7890"));
    }

}

P.S. The regex pattern we use extra '\' for escaping when we use in java string. (Try to use "\d{3}-\d{3}-\d{4}" in java program, you will get an error.

answered Nov 15, 2016 at 5:28

rohan

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Comments

k_g · Accepted Answer · 2014-12-24 05:20:59Z

0

Assuming you want an optimization (which is what your comment suggests).

How bout this? (the "0" is to exclude if they give complete garbage without even a single digit).

 int parse(String phone){
     int num = Integer.parseInt("0"+phone.replaceAll("[^0-9]",""));
     return 100000000<=num&&num<100000000000?num:-1;
 }

answered Dec 24, 2014 at 5:20

k_g

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Comments

Deepak · Accepted Answer · 2016-08-25 00:32:44Z

0

I am not sure but removing the garbage characters parenthesis, spaces and hyphens, if you match with ^((\+[1-9]?[0-9])|0)?[7-9][0-9]{9}$ , you may validate a mobile number

private static final String PHONE_NUMBER_GARBAGE_REGEX = "[()\\s-]+";
private static final String PHONE_NUMBER_REGEX = "^((\\+[1-9]?[0-9])|0)?[7-9][0-9]{9}$";
private static final Pattern PHONE_NUMBER_PATTERN = Pattern.compile(PHONE_NUMBER_REGEX);

public static boolean validatePhoneNumber(String phoneNumber) {
    return phoneNumber != null && PHONE_NUMBER_PATTERN.matcher(phoneNumber.replaceAll(PHONE_NUMBER_GARBAGE_REGEX, "")).matches();
}

answered Aug 25, 2016 at 0:32

Deepak

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Comments

Armer B. · Accepted Answer · 2017-06-21 10:59:39Z

0

One easy and simple to use java phone validation regex:

public static final String PHONE_VERIFICATION = "^[+0-9-\\(\\)\\s]*{6,14}$";

private static Pattern p;
private static Matcher m;

public static void main(String[] args)
{
    //Phone validation
    p = Pattern.compile(PHONE_VERIFICATION);
    m = p.matcher("+1 212-788-8609");
    boolean isPhoneValid = m.matches();

    if(!isPhoneValid)
    {
        System.out.println("The Phone number is NOT valid!");
        return;
    }
    System.out.println("The Phone number is valid!");
}

answered Jun 21, 2017 at 10:59

Armer B.

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Comments

Deepesh Jain · Accepted Answer · 2017-09-20 12:06:13Z

0

i have done testing one regex for this combination of phone numbers

(294) 784-4554
(247) 784 4554
(124)-784 4783 
(124)-784-4783
(124) 784-4783
+1(202)555-0138

THIS REGEX SURELY WILL BE WORKING FOR ALL THE US NUMBERS

\d{10}|(?:\d{3}-){2}\d{4}|\(\d{3}\)\d{3}-?\d{4}|\(\d{3}\)-\d{3}-?\d{4}|\(\d{3}\) \d{3} ?\d{4}|\(\d{3}\)-\d{3} ?\d{4}|\(\d{3}\) \d{3}-?\d{4}

edited Sep 20, 2017 at 12:06

user4447799

answered Sep 20, 2017 at 11:31

Deepesh Jain

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Comments

Jose Martinez · Accepted Answer · 2020-08-15 19:35:15Z

0

Building on @k_g's answers, but for US numbers.

static boolean isValidTelephoneNumber(String number) {
    long num = Long.parseLong("0" + number.replaceAll("[^0-9]", ""));
    return 2000000000L <= num && num < 19999999999L;
}

public static void main(String[] args) {
    var numbers = List.of("+1 212-788-8609", "212-788-8609", "1 212-788-8609", "12127888609", "2127888609",
            "7143788", "736103355");

    numbers.forEach(number -> {
        boolean isPhoneValid = isValidTelephoneNumber(number);
        log.debug(number + " matches = " + isPhoneValid);
    });
}

answered Aug 15, 2020 at 19:35

Jose Martinez

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