将电话号码中的字母和数字转换为所有数字 (Java)

Question

import java.util.Scanner;

import javax.swing.JOptionPane;


public class PhonePadTranslator {

private static Scanner input;

public static void main(String[] args) {

    input = new Scanner(System.in);
    System.out.println("Enter The Phone Number (With Letters) ");
    String initial_phone_number = input.nextLine();
    initial_phone_number = initial_phone_number.toUpperCase();
    int phone_number_final = 0;

    System.out.printf("The phone number for %s is %s", initial_phone_number, phone_number_final);

}//end of main

public static int full_number(String initial_phone_number) 
{
    int which_character = 0;
    int phone_number_final = 0;
    char ch = (Character) null;

    for (which_character = 0; which_character < initial_phone_number.length(); which_character++) 
    {
        if (Character.isLetter(ch)) 
        {
            switch(ch)
            {
        case 'A' : case 'B' : case 'C' : phone_number_final = 2; break;
        case 'D' : case 'E' : case 'F' : phone_number_final = 3; break;
        case 'G' : case 'H' : case 'I' : phone_number_final = 4; break;
        case 'J' : case 'K' : case 'L' : phone_number_final = 5; break;
        case 'M' : case 'N' : case 'O' : phone_number_final = 6; break;
        case 'P' : case 'Q' : case 'R' : case 'S' : phone_number_final = 7; break;
        case 'T' : case 'U' : case 'V' : phone_number_final = 8; break;
        case 'W' : case 'X' : case 'Y' : case 'Z' : phone_number_final =9; break;
            }
            return (char)phone_number_final;
        }
        if (Character.isDigit(ch))
        {
            return (char)phone_number_final;
        }

        else {
            return (char)phone_number_final;
        }

    } //end of for
    return ch;
}//end of full_number
}//end of class

I just thought I'd copy/paste the whole thing... but whenever I run the code, it keeps outputting The phone number for 1800FLOWERS is 0 . Now I'm sure there's some other things that are wrong, but my main concern is why it keeps giving me a 0 ? I feel like it's because I initialised it to that and for some reason I'm never changing the value. Please help, my professor takes forever to respond to my emails :(

Answer 1

Change

int phone_number_final = 0;

to

int phone_number_final = full_number(initial_phone_number);

You did not assign the result to your variable.

Other than that, I believe your full_number function is not exactly correct either.

Updated code:

import java.util.Scanner;

public class StringToNumbers
{
    private static Scanner input;

    public static void main(String[] args)
    {
        input = new Scanner(System.in);
        System.out.println("Enter The Phone Number (With Letters): ");
        String initial_phone_number = input.nextLine();

        initial_phone_number = initial_phone_number.toUpperCase();
        long phone_number_final = full_number(initial_phone_number);

        System.out.printf("%nOutput phone number for '%s' is '%s'",
                initial_phone_number, phone_number_final);
    }

    public static long full_number(String initial_phone_number)
    {
        // Use long instead of int for 'number' if the string will be longer than max int value
        // 2147483647, which is '10 digits'
        long number = 0;
        int strLen = initial_phone_number.length();


        for (int currCharacter = 0; currCharacter < strLen; currCharacter++) 
        {
            char ch = initial_phone_number.charAt(currCharacter);
            // For A-Z & 0-9, multiply by 10, add the 'char' to number.
            // i.e., Shift existing value to the left by 1 digit, add current 'char' to it
            // Use long instead of int if the string will be longer than max int value (2147483647)

            if (Character.isLetter(ch)) 
            {
                switch(ch)
                {
                case 'A' : case 'B' : case 'C' : number *= 10; number += 2; break;
                case 'D' : case 'E' : case 'F' : number *= 10; number += 3; break;
                case 'G' : case 'H' : case 'I' : number *= 10; number += 4; break;
                case 'J' : case 'K' : case 'L' : number *= 10; number += 5; break;
                case 'M' : case 'N' : case 'O' : number *= 10; number += 6; break;
                case 'P' : case 'Q' : case 'R' : case 'S' : number *= 10; number += 7; break;
                case 'T' : case 'U' : case 'V' : number *= 10; number += 8; break;
                case 'W' : case 'X' : case 'Y' : case 'Z' : number *= 10; number += 9; break;
                }
            }
            else if (Character.isDigit(ch))
            {
                number *= 10; number += Character.getNumericValue(ch);
            }
            else
            {
                System.out.println("Invalid character!");
            }

        } // End of for loop

        // Return actual number only at the end of the function
        return number;

    }// End of full_number function    
}

Input/Output:

Enter The Phone Number (With Letters): 
1800FLOWERS

Output phone number for '1800FLOWERS' is '18003569377'

Answer 2

Even though this already has been answered, I'd like to note a few things. Don't use an int or a long to save a phone number! You'll lose leading zeroes! Plus you'll easily go out of your int or long range. Plus the current accepted answer is a little harder to understand. I'd simply go with a lot less, and much easier to understand code:

public String toNormalPhoneNumber(String phoneNumber) {
    String normal = "";
    foreach (char c : phoneNumber.toUppercase().toCharArray())
        normal += getKeypadNumber(c);
    return normal;
}

public char getKeypadNumber(char characterToConvert) {
    if (Character.isDigit(characterToConvert))
        return characterToConvert;
    else {
        switch (characterToConvert) {
            case 'A' : case 'B' : case 'C' : return '2';
            case 'D' : case 'E' : case 'F' : return '3';
            case 'G' : case 'H' : case 'I' : return '4';
            case 'J' : case 'K' : case 'L' : return '5';
            case 'M' : case 'N' : case 'O' : return '6';
            case 'P' : case 'Q' : case 'R' : case 'S' : return '7';
            case 'T' : case 'U' : case 'V' : retrun '8';
            case 'W' : case 'X' : case 'Y' : case 'Z' : return '9';
            default return '?';
        }
    }
}

I reckon this is much easier to understand.

Answer 3

Here is ac# answer

class Program
{
    static void Main(string[] args)
    {
        Console.WriteLine("Please enter the phone number (With Letters): ");
        string initial_phone_number = Console.ReadLine();
        initial_phone_number = initial_phone_number.ToUpper();
        string phone_number_final = full_number(initial_phone_number);

        Console.WriteLine("Output phone number for " + initial_phone_number + " is " + phone_number_final);
        Console.ReadLine();
    }

    public static string full_number(String initial_phone_number)
    {
        string number = "";
        string digit = "";
        int strLen = initial_phone_number.Length;

        for (int currCharacter = 0; currCharacter < strLen; currCharacter++)
        {
            string ch = initial_phone_number.Substring(currCharacter,1);
            int n;
            bool isNumeric = int.TryParse(ch, out n);
            if (!isNumeric)
            {
                switch (ch)
                {
                    case "A": case "B": case "C": digit = "2"; break;
                    case "D": case "E": case "F": digit = "3"; break;
                    case "G": case "H": case "I": digit = "4"; break;
                    case "J": case "K": case "L": digit = "5"; break;
                    case "M": case "N": case "O": digit = "6"; break;
                    case "P": case "Q": case "R": case "S": digit = "7";  break;
                    case "T": case "U": case "V": digit = "8";  break;
                    case "W": case "X": case "Y": case "Z": digit = "9";  break;

                }
                number = number + digit;
            }
            else if (isNumeric)
            {
                number = number + n.ToString();
            }
            else
            {
                Console.WriteLine("Invalid character!");
            }

        } 

        return number;

    }

}