How to parse 10 digit phone number into Integer in Java

Question

I have to do a validation where it is mentioned that phone number should be 10 digits like 9867610012 and if its true I have to set its value in a setter method whose argument is an Integer and I can't change it's argument. As we know, Integer has a minimum value of -2,147,483,648 and a maximum value of 2,147,483,647 (inclusive). When I parse the String to Integer using Integer.parseInt("9867610012"), I am getting following exception at runtime:

Exception in thread "main" java.lang.NumberFormatException: For input string: "9867610012".

Can you suggest some way where I can set a 10 digit String to Integer.

Thanks in Advance!!!!

Thats a very bad idea btw. A phone number should never be an Integer. If you have a 0 as prefix (which is very common) it will just be removed. An international number starts with something like 0049123123 ... parsing that into an integer will give you a broken number — djnose
– djnose, Commented Sep 24, 2017 at 6:39
Don't fall into the trap to think a phone number can be represented as an int or long. It is a string of digits, not a number. Remember, phone numbers can start with 0. — Henry
– Henry, Commented Sep 24, 2017 at 6:39
Why? Phone numbers aren't integers. You should treat them as indicitave data, not as values. — user207421
– user207421, Commented Sep 24, 2017 at 6:42

Mike B · Accepted Answer · 2017-09-24 07:25:16Z

1

As I commented, You should leave it as a String and then validate using regex.
In Your situation I'll give an example (10 digits, may start with zero). Don't assume that my code will be perfect - it's just to give you an idea.

boolean isValid = false;
do {
   Scanner phoneSc = new Scanner(System.in);
   String phone = phoneSc.next();
   isValid = isPhoneValid(phone);
} while (!isValid);


public boolean isPhoneValid(String phone) {
    String regex = "\\d{10}"; //regex for 10 digits
    return phone.matches(regex);
}

If You don't know regex, learn them. They are very useful. If You don't want to learn, web is full of examples. You can add to regex possible + in front, ( and ), maybe some dashes - cause some numbers have it. Regexes are perfect for such task.

answered Sep 24, 2017 at 7:25

Mike B

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Comments

Kaushik NP · Accepted Answer · 2017-09-24 07:21:36Z

0

Use parseLong().

public static long parseLong(String s) throws NumberFormatException

Parses the string argument as a signed decimal long. The characters in the string must all be decimal digits, except that the first character may be an ASCII minus sign '-' (\u002D') to indicate a negative value or an ASCII plus sign '+' ('\u002B') to indicate a positive value.

Parameters: s - a String containing the long representation to be parsed

Returns: the long represented by the argument in decimal.

Throws: NumberFormatException - if the string does not contain a parsable long.

And since Long.MAX_VALUE is 9,223,372,036,854,775,807, it can serve your purpose. So instead of Integer use Long.

Note to OP : as many have pointed out, you may need to review your code logic to consider various scenarios that you may come across later on. Do go through the comments once.

edited Sep 24, 2017 at 7:21

answered Sep 24, 2017 at 6:41

Kaushik NP

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4 Comments

Mike B Over a year ago

hey man. read the comments plz. it's not a good idea to parse phone numbers as int or long.

Kaushik NP Over a year ago

The comments do mention that, but right now the OP has not mentioned them. His title mentions "How to parse 10 digit phone number into Integer in Java" and even his question is directed at that.

Kaushik NP Over a year ago

I agree that he might need to work out the logic again for better coverage of cases, but thats for him to work on. I am just helping him with his error and clarifying what he can do to avoid it.

OneCricketeer Over a year ago

This answer doesn't address leading zeros in the numbers

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