将电话号码中的字母和数字转换为所有数字 (Java)
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Convert Letters & Numbers in a Phone Number to all Numbers (Java)
提问by John Sideris
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 :(
我只是想我会复制/粘贴整个内容......但是每当我运行代码时，它都会不断输出The phone number for 1800FLOWERS is 0。现在我确定还有其他一些事情是错误的，但我主要关心的是为什么它一直给我一个0? 我觉得这是因为我将其初始化为该值，并且出于某种原因我从未更改过该值。请帮忙，我的教授需要永远回复我的电子邮件:(
采纳答案by almightyGOSU
Change
改变
int phone_number_final = 0;
to
到
int phone_number_final = full_number(initial_phone_number);
You did not assign the result to your variable.
您没有将结果分配给您的变量。
Other than that, I believe your full_numberfunction is not exactly correct either.
除此之外，我相信您的full_number功能也不完全正确。
Updated code:
更新代码：
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:
输入输出：
Enter The Phone Number (With Letters):
1800FLOWERS
Output phone number for '1800FLOWERS' is '18003569377'
回答by stealthjong
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:
尽管已经回答了这个问题，但我想指出一些事情。不要使用 int 或 long 来保存电话号码！您将丢失前导零！另外，你很容易超出你的整数或远程范围。另外，目前接受的答案有点难以理解。我只会使用更少且更容易理解的代码：
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.
我认为这更容易理解。
回答by pat capozzi
Here is a c# answer
这是 ac# 的答案
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;
}
}
