扫描器内方法不等待应答

Question

已经纠结了一天了，看了论坛的讨论来来回回，没有结果。谁能告诉我为什么函数aMenu()的第二次调用返回零，而不是等待新的用户输入?我尝试了各种各样的东西，比如hasNextInt()， nextLine()，没有工作。不应该有nextint()块，直到用户写东西?我怎么解决这个问题?谢谢。

package FirstJavaPackage;
import java.util.Scanner;
public class testScanner
{
    public static void main(String[] args)
    {
        int choice = aMenu();
        System.out.println("You typed: "+choice);
        choice = aMenu();
        System.out.println("You typed: "+choice);
    }
    public static int aMenu()
    {
        int result = 0;
        System.out.println("In aMenu... enter an int: ");
        Scanner keyboard = new Scanner(System.in);
        if (keyboard.hasNextInt())
            result = keyboard.nextInt();
        keyboard.close();
        return result;
    }
}

输出为:

在aMenu

…输入一个int:2你输入:2在aMenu…输入一个int:您键入:0

Answer 1

您需要在调用aMenu()时重用相同的Scanner对象:

public static void main(String[] args)
{
    Scanner keyboard = new Scanner(System.in);
    int choice = aMenu(keyboard);
    System.out.println("You typed: "+choice);
    choice = aMenu(keyboard);
    System.out.println("You typed: "+choice);
}
public static int aMenu(Scanner keyboard)
{
    int result = 0;
    System.out.println("In aMenu... enter an int: ");
    result = keyboard.nextInt();
    return result;
}

有关进一步的讨论，请参见如何在System.in上使用多个Scanner对象?

Answer 2

在第一次调用之后，您实际上关闭了System.in输入流。

来自Scanner.close()文档:

When a Scanner is closed, it will close its input source if the source 
implements the Closeable interface.

尽量不要在aMenu的末尾close您的扫描仪:在aMenu方法之外初始化扫描仪，并使该方法使用它。

Answer 3

由于scanner.close将关闭整个输入源，您应该将scanner传递给您的aMenu方法并执行如下操作:

import java.util.Scanner;
public class TestScanner
{
   public static void main(String[] args)
   {
      Scanner keyboard = new Scanner(System.in);
      int choice = 0;
      do
      {
         choice = aMenu(keyboard);
         System.out.println("You typed: " + choice);
      } while (choice > 0);
      keyboard.close();
    }
    public static int aMenu(Scanner keyboard)
    {
        int result = 0;
        System.out.println("In aMenu... enter an int: ");
        if (keyboard.hasNextInt())
           result = keyboard.nextInt();
        return result;
    }
}