控制台忽略输入

Question

我创建了一个程序，使用Scanner接收用户的数字，并将其保存到从1到100之间的整数“a”。请参阅下面的Java文件：

public class Parity_Check {
  private static Scanner sc;

  public static void main(String[] args) throws InterruptedException {

    sc = new Scanner(System.in);
    int a, b;
    System.out.print("Enter a number     between 1 and 100: ");

    while(true) {
      b = 0;
      if(!sc.hasNextInt()) {
        System.out.print("That isn't an integer! Try again: "); 
        sc.next();
      }
      else{
        b = sc.nextInt();
        if(b < 1 || b > 100) {
          System.out.print("That integer isn't between 1 and 100! Try again: "); 
          sc.next();
        }
        else{
          a = b; 
          break;
        }
      }
    }
    System.out.print("The number is: "+a+".");
  }
}

我遇到的问题如下:在程序返回“该整数不是介于1和100之间！再试一次：“它等待来自用户的两个输入(而不是它应该输入的输入)--第一个输入被完全忽略！下面是我为说明这个问题而运行的控制台会话：

"Enter a number between 1 and 100: 2.5
That isn't an integer! Try again: 101
That integer isn't between 1 and 100! Try again: Apple.
42
The number is: 42.”

正如您所看到的，它甚至没有注意到输入"Apple".，我完全不明白为什么它不能正常工作，如下所示：

"Enter a number between 1 and 100: 2.5
That isn't an integer! Try again: 101
That integer isn't between 1 and 100! Try again: Apple.
That isn't an integer! Try again: 42
The number is: 42.”

我对Java非常陌生，所以一个解释得很好的答案是“上帝之神”；我更感兴趣的是为什么它不能工作，而不是如何修复它，因为希望我能够学到它。

顺便说一下，我使用的是最新版本的Eclipse 64位。

Answer 1

在这里，您已经从流中删除了一个int，所以您必须删除更多的内容。发出对sc.next()的调用：

    b = sc.nextInt();
    if(b < 1 || b > 100) {
      System.out.print("That integer isn't between 1 and 100! Try again: "); 
      // sc.next(); remove this
    }

注意，这与前面的if -语句的情况是如何不同的:如果用户输入的东西不是一个数字，您必须通过调用next()将其从流中删除，因为否则没有其他任何东西会删除它。在这里，对nextInt的调用已经从流中删除了输入。

Answer 2

您的if语句中的sc.next()是不需要的，它使您的代码跳过用户的下一个输入。下面的代码片段按照您的意愿正确工作。

private static Scanner sc;

public static void main(String[] args) throws InterruptedException {

    sc = new Scanner(System.in);
    int a, b;
    System.out.println("Enter a number     between 1 and 100: ");

    while(true) {
        b = 0;
        if(!sc.hasNextInt()) {
            System.out.println("That isn't an integer! Try again: ");
            sc.next(); 
        }
        else {
            b = sc.nextInt();
        }
        if(b < 1 || b > 100) {
            System.out.println("That integer isn't between 1 and 100! Try again: "); 
        }
        else {
            a = b; 
            break;
        }
    }
    System.out.println("The number is: "+a+".");
    return;
}

Answer 3

试试下面的主板，它会起作用的

公共静态空主(String[] args)抛出InterruptedException {

sc = new Scanner(System.in);
int a, b;
System.out.print("Enter a number     between 1 and 100: ");

while(true) {
  b = 0;
  if(!sc.hasNextInt()) {
    System.out.print("That isn't an integer! Try again: "); 
    sc.next();
  }
  else{
    b = sc.nextInt();
    if(b < 1 || b > 100) {
      System.out.print("That integer isn't between 1 and 100! Try again: "); 
    }
    else{
      a = b; 
      break;
    }
  }
}
System.out.print("The number is: "+a+".");

}