how count the number of digits of telephone numbers?

Question

easy for usual numbers, but telephone numbers can start with 01-...., as an example, 01234 is basically 1234 for java, right?

So I can't divide by 10 recursively to find out how many digits there are. is there any different way to find out how many digits there are?

thanks in advance. ps.: no regex if possible

Answer 1

Assume that the phone number is a string,

String pn = "049-4912394129" // that's a random value

then you could iterate in that string and check if a character is indeed a number

int count = 0;
for(char c : pn.toCharArray()){
  if(Character.isDigit(c))
    count++;
}

Answer 2

As you phone number is an int, you don't need to bother with regex and every locale phone number patterns.

Simply convert your int phone number to a String. Then you can easily get the string length.

int myIntNumber = 1234;
String myStringNumber = String.valueOf(myIntNumber);

int length = myStringNumber.length();

Answer 3

this could easily be done with a lambda

"049-4912394129".codePoints().filter( Character::isDigit ).count();  // 13

Answer 4

If you are talking about a number that represented by a String object then:

public int getNumberOfDigits(phoneNumber) {
   int count = 0;
   for (int i = 0, i < phoneNumber.length(); i++) {
       if (Character.isDigit(phoneNumber.charAt(i))) {
           count++;
       }
   }
   System.out.println("Number of digits: " + count);
   return count;
}

If you are talking about a number that represented by an int then simply convert it to String before using it like so:

String phoneNumber = String.valueOf(phoneNumberInInt);  

I assumed you DO want to count the zeros as a digit because you are not summarizing the value of them, so they do have a meaning when you talk about how many digits are there.