将字母映射到电话键盘上的数字

Question

我似乎找不到这段代码的问题。我正在尝试将由数字和字母组成的电话号码仅转换为数字。例如，1800SMILEING 应转换为 1800-764-5464。但是我的代码重复了每组中的最后一个数字。 180048-766-5466 而不是正确的格式。它还会在 800 之后生成额外的 48。 请帮忙，我已经为我的 Java 课程作业做了很多小时，但我无法找出问题所在。

import java.util.Scanner;

public class PhoneNumberConverter {

    public static void main(String[] args) {
        System.out.println("Enter a phone number to convert:");
        Scanner input = new Scanner(System.in);
        String phoneNumber = input.next();
        
        input.close();
        
        int firstGroup = translatePhoneNumber (phoneNumber, 0, 4);
        System.out.print(firstGroup);
        System.out.print("-");
        
        int secondGroup = translatePhoneNumber (phoneNumber, 4, 6);
        System.out.print(secondGroup);
        System.out.print("-");
        
        int thirdGroup = translatePhoneNumber (phoneNumber, 7, 10);
        System.out.print(thirdGroup);
        
    }
    
    public static int translatePhoneNumber (String phoneNumber, int firstIndex, int lastIndex) {
    
        int chartoNumber = 'A';
        int currentIndex;   
        if (firstIndex != 0) {
        
            for (currentIndex = firstIndex; currentIndex < lastIndex; currentIndex++) {
                if (phoneNumber.charAt(currentIndex) == 'A' || phoneNumber.charAt(currentIndex) == 'B' || phoneNumber.charAt(currentIndex) == 'C' )
                    chartoNumber = 2;
                else if (phoneNumber.charAt(currentIndex) == 'D' || phoneNumber.charAt(currentIndex) == 'E' || phoneNumber.charAt(currentIndex) == 'F' )
                    chartoNumber = 3;
                else if (phoneNumber.charAt(currentIndex) == 'G' || phoneNumber.charAt(currentIndex) == 'H' || phoneNumber.charAt(currentIndex) == 'I' )
                    chartoNumber = 4;
                else if (phoneNumber.charAt(currentIndex) == 'J' || phoneNumber.charAt(currentIndex) == 'K' || phoneNumber.charAt(currentIndex) == 'L' )
                    chartoNumber = 5;
                else if (phoneNumber.charAt(currentIndex) == 'M' || phoneNumber.charAt(currentIndex) == 'N' || phoneNumber.charAt(currentIndex) == 'O' )
                    chartoNumber = 6;
                else if (phoneNumber.charAt(currentIndex) == 'P' || phoneNumber.charAt(currentIndex) == 'Q' || phoneNumber.charAt(currentIndex) == 'R' || phoneNumber.charAt(currentIndex) == 'S' )
                    chartoNumber = 7;
                else if (phoneNumber.charAt(currentIndex) == 'T' || phoneNumber.charAt(currentIndex) == 'U' || phoneNumber.charAt(currentIndex) == 'V' )
                    chartoNumber = 8;
                else if (phoneNumber.charAt(currentIndex) == 'W' || phoneNumber.charAt(currentIndex) == 'X' || phoneNumber.charAt(currentIndex) == 'Y' || phoneNumber.charAt(currentIndex) == 'Z' )
                    chartoNumber = 9;
                else
                    chartoNumber =  phoneNumber.charAt(currentIndex);
                
                    System.out.print(chartoNumber); 
                    
                }
        } else {
            for (currentIndex = firstIndex; currentIndex < lastIndex; currentIndex++) {
                chartoNumber =  phoneNumber.charAt(currentIndex);
                char numbeeer = (char) chartoNumber;
                System.out.print(numbeeer);
            }   
        }
        return chartoNumber;    
    }       
}

Answer 1

首先请注意，您是在translatePhoneNumber 方法本身中打印出结果，您不需要再次打印出结果。

摆脱System.out.print(firstGroup);等

第二，if (firstIndex != 0) { - 为什么？请注意，您正在执行 translatePhoneNumber (phoneNumber, 0, 4); 所以 firstIndex 将是 0

我个人只会将子字符串传递给方法并为每个字符循环。

translatePhoneNumber (phoneNumber.substring (0, 4));

....

for (char c : localPhoneNumber) {
    // your mapping
}