如何使用循環獲取全名字符串的第一個字母？

Question

編寫一個應用程序，提示用戶輸入全名和街道地址，並根據用戶姓名的首字母和數字部分構造一個ID。例如，居住在34 Elm的用戶William Henry Harrison將具有WHH34的ID，而住在1778門羅的用戶Addison Mitchell將具有AM1778的ID。將該文件保存爲ConstructID.java。

這是迄今爲止，W/C我的老師說，它的錯......

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：我不能使用數組 :(這個問題的話題是關於循環和字符串..沒有陣列..所以請幫助.. :(（

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

如果你不能使用數組，請使用列表。查看javadocs的java.util.List以瞭解列表可以做什麼。講座筆記！）

我期望這個理由你的老師說過你寫的是錯的是它假設爲大家的名字由名字，中間名和姓氏組成。我相信你知道這是不正確的。有些人有許多中間名，或根本沒有。的確，有些人只有一個名字。

你的老師想要的是能夠應對任意名字的代碼。一個數組將是一個不好的選擇...因爲你需要預測在之前你從數組中讀取數組的大小，這個數組是。

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);