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这个程序是由 3 个数字和 7 个字母组成的字符串，使用方法将其转换为等效的数字。用户输入是一个字符串，它从字母变为数字，然后再变成由破折号分隔的数字字符串。所以这只是学习的第 3 周，我被方法绊倒了。我收到的错误代码告诉我字符串索引超出范围。所以，我怀疑错误出在我的转换中，或者我在第 89 行附近连接了最后的字符串。我花了 20 分钟与 TA 交流，希望得到任何指导。我仍然与该程序有更多的关系，所以我省略了一些，因为我只需要在我遇到卡住超过 45 分钟的地方时获得帮助。主要问题 - 错误发生在哪里，为什么？我试图将打印语句放在任何地方，看看发生了什么......

import java.util.Scanner;
public class ConvPhoneNumTwo{
  public static void main(String[] args){
    String s = getInput();


    printPhone(convertToDigit(s));
  }   

 public static String getInput() { 
      Scanner scan = new Scanner(System.in);
      String s1 = "";
      boolean length = false;
      while (length==false){ 
        System.out.println(" Please enter a telephone number in this format ###-AAA-AAAA") ;
         s1 = scan.nextLine();
        if(s1.length() != 12 ){
          System.out.println( "this is an invalid choice try again Plz..."); 
           length = false;

        }
        else{ 
          length = true; 
        }
      }
      return s1; 
    }

 public static String convertToDigit(String s010){
    s010 = s010.toLowerCase();
    String s001= s010.substring(0,3);
    String s002 = s010.substring(4,6);
    String s003 = s010.substring(8,12);
    String s1 = (s001+s002+s003);
    String s2 = "";

    // Exceptions to our problem to stop invalid inputs 
    if (s1 == null) {
      System.out.print (" invalid input null thing"); 
    }
    if (s1.length() != 10){
      System.out.print (" invalid input"); 
    }

    String s6 = "";
    for (int i=0; i < s1.length(); i++) {


      if (Character.isDigit(s1.charAt(i))){
        s2 += s1.charAt(i); 
      }

      //sorting of the letter inputs 
      char ch = s1.charAt(i); 

      if (ch == 'a'||ch=='b'||ch== 'c'){
        s2 += "2"; 
      }
      if (ch == 'd'||ch=='e'||ch=='f'){
        s2 += "3"; 
      }
      if (ch == 'g'||ch=='h'||ch=='i'){
        s2 += "4"; 
      }
      if (ch == 'j'||ch=='k'||ch=='l'){
        s2 += "5"; 
      }
      if (ch == 'm'||ch=='n'||ch=='o'){
        s2 += "6"; 
      }
      if (ch == 'p'||ch=='q'||ch=='r'|| ch=='s'){
        s2 += "7"; 
      }
      if (ch == 't'||ch=='u'||ch=='v'){
        s2 += "8"; 
      }
      if (ch == 'w'||ch=='x'||ch=='y'|| ch=='z')
      {
        s2 += "9"; 
      }
      else{ 
      }

     String s3 = s2.substring(0,3);
    String s4 = s2.substring(3,6);
    String s5 = s2.substring(6);
     s6 = ( s3 +"-"+ s4 + "-"+ s5);
    }

    return s6; 
    }

    public static void printPhone(String message) { //print out whatever string it is given, basically System.out.println(), but through a method instead
    System.out.println(message);
  }
}

不确定我可以在其中添加多少细节。

4

1 回答 1

3

我想我找到了你的问题。

在将输入转换为数字的方法中，您这样写：

String s6 = "";
for (int i=0; i < s1.length(); i++) {

  if (Character.isDigit(s1.charAt(i))){
    s2 += s1.charAt(i); 
  }

  //sorting of the letter inputs 
  char ch = s1.charAt(i); 

  if (ch == 'a'||ch=='b'||ch== 'c'){
    s2 += "2"; 
  }
  if (ch == 'd'||ch=='e'||ch=='f'){
    s2 += "3"; 
  }
  if (ch == 'g'||ch=='h'||ch=='i'){
    s2 += "4"; 
  }
  if (ch == 'j'||ch=='k'||ch=='l'){
    s2 += "5"; 
  }
  if (ch == 'm'||ch=='n'||ch=='o'){
    s2 += "6"; 
  }
  if (ch == 'p'||ch=='q'||ch=='r'|| ch=='s'){
    s2 += "7"; 
  }
  if (ch == 't'||ch=='u'||ch=='v'){
    s2 += "8"; 
  }
  if (ch == 'w'||ch=='x'||ch=='y'|| ch=='z')
  {
    s2 += "9"; 
  }
  else{ 
  }

 String s3 = s2.substring(0,3);
 String s4 = s2.substring(3,6);
 String s5 = s2.substring(6);
 s6 = ( s3 +"-"+ s4 + "-"+ s5);
}

首先，您检查了输入字符串中是否已经有任何数字 - 如果有，那么您只需将数字添加到已处理的字符串中。然后您检查字母并相应地处理每个字母。

但是你有一个问题。在完成已处理字符串的构造s2（在for循环内）之前，您尝试提取已处理字符串的部分内容。这应该在for循环之后完成。

因此，该方法应该看起来像这样。

 public static String convertToDigit(String s010){
    s010 = s010.toLowerCase();
    String s001= s010.substring(0,3);
    String s002 = s010.substring(4,6);
    String s003 = s010.substring(8,12);
    String s1 = (s001+s002+s003);
    String s2 = "";

    // Exceptions to our problem to stop invalid inputs 
    if (s1 == null) {
      System.out.print (" invalid input null thing"); 
    }
    if (s1.length() != 10){
      System.out.print (" invalid input"); 
    }

    String s6 = "";
    for (int i=0; i < s1.length(); i++) {


      if (Character.isDigit(s1.charAt(i))){
        s2 += s1.charAt(i); 
      }

      //sorting of the letter inputs 
      char ch = s1.charAt(i); 

      if (ch == 'a'||ch=='b'||ch== 'c'){
        s2 += "2"; 
      }
      if (ch == 'd'||ch=='e'||ch=='f'){
        s2 += "3"; 
      }
      if (ch == 'g'||ch=='h'||ch=='i'){
        s2 += "4"; 
      }
      if (ch == 'j'||ch=='k'||ch=='l'){
        s2 += "5"; 
      }
      if (ch == 'm'||ch=='n'||ch=='o'){
        s2 += "6"; 
      }
      if (ch == 'p'||ch=='q'||ch=='r'|| ch=='s'){
        s2 += "7"; 
      }
      if (ch == 't'||ch=='u'||ch=='v'){
        s2 += "8"; 
      }
      if (ch == 'w'||ch=='x'||ch=='y'|| ch=='z')
      {
        s2 += "9"; 
      }
      else{ 
      }

// They should not be here
/*   String s3 = s2.substring(0,3);
    String s4 = s2.substring(3,6);
    String s5 = s2.substring(6);
     s6 = ( s3 +"-"+ s4 + "-"+ s5);*/
    } // end of for loop; completed constructing s2

// They should be here
String s3 = s2.substring(0,3);
String s4 = s2.substring(3,6);
String s5 = s2.substring(6);
s6 = ( s3 +"-"+ s4 + "-"+ s5);

return s6; 
}
于 2017-03-08T03:44:12.940 回答