如何使用循环获取全名字符串的第一个字母？

Question

编写一个应用程序，提示用户输入全名和街道地址，并根据用户姓名的首字母和数字部分构造一个ID。例如，居住在34 Elm的用户William Henry Harrison将具有WHH34的ID，而住在1778门罗的用户Addison Mitchell将具有AM1778的ID。将该文件保存为ConstructID.java。

这是迄今为止，W/C我的老师说，它的错......

import java.util.*; 
public class ConstructID { 

    public static void main(String[] args) { 

     String name1, address1, address2; 

     Scanner kevs = new Scanner(System.in); 

     System.out.println("Enter your fullname. Encluding Middle Initial, Separated by spaces."); 

     name1 = kevs.next(); 
     name2 = kevs.next(); 
     name3 = kevs.next(); 

     name1 = name1.toUpperCase(); 
     name2 = name2.toUpperCase(); 
     name3 = name3.toUpperCase(); 

     name1 = name1.substring(0,1); 
     name2 = name2.substring(0,1); 
     name3 = name3.substring(0,1); 

     System.out.println("\nEnter your address. Separated by spaces."); 

     address1 = kevs.next(); 
     address2 = kevs.nextLine(); 


     do { 

      if (address1 == address1.substring(0,1) || address1 == address1.substring(0,2) || address1 == address1.substring(0,3) || address1 == address1.substring(0,4) || address1 == address1.substring(0,5) || address1 == address1.substring(0,6)) 
      System.out.println("\nYour ID: " + name1 + name2 + name3 + address1); 

     } while (address1 == address2); 
    } 
} 

注：我不能使用数组 :(这个问题的话题是关于循环和字符串..没有阵列..所以请帮助.. :(（

Answer 1

public static void main(String[] args) { 

     String name, address; 

     Scanner kevs = new Scanner(System.in); 

     System.out.println("Enter your fullname. Including Middle Initial, Separated by spaces."); 

     //get the full name 
     name = kevs.nextLine(); 

     System.out.println("\nEnter your address. Separated by spaces."); 
     //get the address 
     address = kevs.nextLine(); 

     String initials = ""; 
     //get the first letter of the name and add it to our initial string 
     char c = name.charAt(0); 
     initials += c; 
     for (int i = 0; i < name.length(); i++) { 
      char letter = name.charAt(i); 
      // if we find a space, select the first letter after it until the end 
      if (letter == ' ') { 
       initials += name.charAt(i + 1); 
      } 

     } 
     String addressNum = ""; 
     //this bool is so that we only select characters up to the first space 
     boolean finished = false; 
     for (int i = 0; i < address.length(); i++) { 
      if (!finished) { 
       char num = address.charAt(i); 
       if (num != ' ') { 
        //add characters to the address string until there is a space 
        addressNum += num; 
       } else // we found the first space so we are now finished 
       { 
        finished = true; 
       } 
      } else //we are finished so leave the loop 
      { 
       break; 
      } 
     } 
     //concatenate the strings 
     System.out.println(initials + addressNum); 

    } 

Answer 2

如果你不能使用数组，请使用列表。查看javadocs的java.util.List以了解列表可以做什么。讲座笔记！）

我期望这个理由你的老师说过你写的是错的是它假设为大家的名字由名字，中间名和姓氏组成。我相信你知道这是不正确的。有些人有许多中间名，或根本没有。的确，有些人只有一个名字。

你的老师想要的是能够应对任意名字的代码。一个数组将是一个不好的选择...因为你需要预测在之前你从数组中读取数组的大小，这个数组是。

Answer 3

String name1, address1; 

    Scanner kevs = new Scanner(System.in); 

    System.out.println("Enter your fullname. Encluding Middle Initial, Separated by spaces.\n"); 
    name1 = kevs.nextLine(); 
    name1 = name1.toUpperCase(); 
    StringTokenizer tokens = new StringTokenizer(name1); 
    String name = ""; 
    while(tokens.hasMoreTokens()) { 
     String value= tokens.nextToken(); 
     name += value.substring(0,1); 
    } 
    System.out.println("\nEnter your full address. Separated by spaces.\n"); 
    address1 = kevs.nextLine(); 
    address1 = address1.toUpperCase(); 
    StringTokenizer tokens2 = new StringTokenizer(address1); 
    Integer numericAddress = null; 
    while(tokens2.hasMoreTokens()) { 
     String value1= tokens2.nextToken(); 
     try { 
      numericAddress = Integer.valueOf(value1); 
     }catch(NumberFormatException ne) { 
      continue; 
     } 
     break; 
    } 

    String output = name+numericAddress.toString(); 

    System.out.println(output);