Java Homework - 使用next()和nextLine()方法的Scanner对象

时间:2012-11-16 04:26:04

标签: java java.util.scanner

我正在编写一个main方法来为我正在为我正在学习的java类编写的Contact类和ContactBook类提供一个菜单。我的问题是,我希望当用户输入A,F,P或Q时,我的扫描仪对象(kbd)将捕获输入,使用它,并在输入下一个输入后继续。显然有一些关键是我不理解,因为推动回归并不总能像我预期的那样推进我的计划。我已经包含了我的代码和输出。任何提示将不胜感激。

  import java.util.Scanner;
  public class run{
  public static void main(String[]args){
      Scanner kbd = new Scanner(System.in);
      boolean quit = false;
      System.out.println("How many contacts would you like in your Contact Book?: ");
      int size = kbd.nextInt();
      kbd.nextLine();
      ContactBook kevin = new ContactBook(size);
      while(!quit){
         System.out.println("A - Add a contact \n"+
                            "F - Find a contact \n"+
                            "P - Prints the list \n"+
                            "Q - Quits");


         if(kbd.next().equals("A")){
            if(ContactBook.full(kevin))
               System.out.println("Contact book full!");
            else{
               Contact temp = new Contact();
               System.out.println("Enter a First Name: ");
               temp.setFirstName(kbd.nextLine());
               System.out.println("Enter a Last Name: ");
               temp.setLastName(kbd.nextLine());
               System.out.println("Enter a Phone Number: ");
               temp.setPhoneNumber(kbd.nextLine());
               System.out.println("Enter an email: ");
               temp.setEmail(kbd.nextLine());
               kevin.addContact(temp);
            }
         }
         if(kbd.next().equals("F")){
            kevin.search();
         }
         if(kbd.next().equals("P")){
            System.out.print(kevin.produce());
         }
         if(kbd.next().equals("Q")){
            quit = true;
         }
      }
   }
}

这是我得到的输出。

 ----jGRASP exec: java run
How many contacts would you like in your Contact Book?: 
3
A - Add a contact 
F - Find a contact 
P - Prints the list 
Q - Quits
A
Enter a First Name: 
Kevin
Enter a Last Name: 
Smith
Enter a Phone Number: 
312-4567
Enter an email: 
kevin@gmail.com



 //here I keep pushing enter and am not sure why it doesn't continue back to
 //the beginning of my while loop



a
a
a
A - Add a contact 
F - Find a contact 
P - Prints the list 
Q - Quits

a

 ----jGRASP: process ended by user.

 ----jGRASP exec: java run
How many contacts would you like in your Contact Book?: 

 ----jGRASP: process ended by user.

 ----jGRASP exec: java run
How many contacts would you like in your Contact Book?: 
4
A - Add a contact 
F - Find a contact 
P - Prints the list 
Q - Quits
A
Enter a First Name: 
Enter a Last Name: 
Smith
Enter a Phone Number: 
312-4567
Enter an email: 
kevin@gmail.com

a
s
d
A - Add a contact 
F - Find a contact 
P - Prints the list 
Q - Quits

同样,我是一名学生,这是我的第二个Java课程。我检查了很多资源,试图了解我做错了什么,而且我无法将它拼凑在一起。希望有人可以为我阐明这一点。感谢。

1 个答案:

答案 0 :(得分:3)

我会删除对kbd.next()的所有来电并替换kbd.nextLine()。这里没有必要使用next(),因为它不处理行结束令牌,所以它会让你感到困惑。如果您绝对需要使用kbd.next(),请务必在致电kbd.nextLine()后致电next(),以允许您的程序以合理的方式处理行结束令牌。