仅限严格整数如果/然后循环

Question

使用Java编译接受任何输入的代码，从字符串到双精度，浮点数，整数等，但只能基于整数进行处理。如果它收到一个double，float或者一个字符串，它只会产生一条提示用户再次尝试的消息。我不确定如何滚动，所以这是我现在的代码。

import java.util.Scanner; 

public class Sand 
{ 
    public static void main(String[] args) 
    { 
    int firstInt, secondInt, thirdInt; 

    Scanner keyboard = new Scanner(System.in); 
    System.out.println("Hi. I'm going to be asking you for three integers in just a moment."); 
    System.out.println("An integer is defined as a WHOLE number that's not a fraction or decimal."); 
    System.out.println("I swear to you, I will go off if you put in a non-whole number..."); 
    System.out.println("Okay go ahead and type in the first of the three integers:"); 

    if (keyboard.hasNextInt()) 
    { 
     firstInt = keyboard.nextInt(); 
     System.out.println("Red."); 
     System.out.println("So the first integer is " + firstInt); 
     System.out.println("Okay go ahead and type in the second of the three integers: "); 
     if (keyboard.hasNextInt()) 
     { 
      secondInt = keyboard.nextInt(); 
      System.out.println("Green."); 
      System.out.println("So the first and second integers are " + firstInt + " and " + secondInt); 
     } 

     else 
     { 
      System.out.println("Orange."); 
      System.out.println("Please try again."); 
     } 
    } 
    else 
    { 
     System.out.println("Blue."); 
     System.out.println("Please try again."); 
    } 
    } 
} 

我有橙色和蓝色表示有一些输入错误，但它不完整的时间。我不知道如何处理这个问题，无论是循环还是for循环或try/catch。在学习Java时，我是新手，所以一些＃注解在这方面会很有帮助。该代码旨在从字符串中读取用户中的三个整数。这很简单，但分析输入是我挣扎的地方。

Answer 1

本应该做的工作..

import java.util.Scanner; 

public class Main { 

public static void main(String[] args) { 

    int[] intArray = new int[3]; 
    Scanner keyboard = new Scanner(System.in); 

    for (int i = 0; i < intArray.length; i++) { 

     System.out.print("input integer #"+(i+1)+": "); 
     if (keyboard.hasNextInt()) { 
      intArray[i] = keyboard.nextInt(); 
      System.out.println("value for #"+(i+1)+": " + intArray[i]); 
     } else 
     { 
      System.out.println("not an integer"); 
      break; 
     } 

    } 

    keyboard.close(); 

} 

} 

也许是再次询问时数没有整数

import java.util.Scanner; 

public class Main2 { 

public static void main(String[] args) { 

    int[] intArray = new int[3]; 
    Scanner keyboard = new Scanner(System.in); 

    for (int i = 0; i < intArray.length; i++) { 

     boolean isInputCorrect = false; 
     do { 
      System.out.print("input integer #" + (i + 1) + ": "); 
      if (keyboard.hasNextInt()) { 
       intArray[i] = keyboard.nextInt(); 
       System.out.println("value for #" + (i + 1) + ": " + intArray[i]); 
       isInputCorrect = true; 
      } else { 
       System.out.println("not an integer, try again"); 
       keyboard.nextLine(); 
      } 

     } while (!isInputCorrect); 

    } 

    keyboard.close(); 

} 

} 

Answer 2

这并不意味着完全而是点回答你的问题的另一个解决方案你在正确的方向。这里应该有足够的帮助你弄清楚你的程序还需要什么。

import java.util.Scanner; 

public class Sandbox { 
    // arguments are passed using the text field below this editor 
    public static void main(String[] args) { 
    Scanner keyboard = new Scanner(System.in); 
    int num = 0; 
    System.out.println("enter an integer"); 

    while (true) { //keep prompting the user until they comply 
     try { 
      num = Integer.parseInt(keyboard.nextLine()); 
      keyboard.close(); 
      break; 
     } catch (NumberFormatException e) { 
      System.out.println("You must enter an integer"); 
     } 
    } 
    System.out.println("num: " + num); 
    } 
} 

Answer 3

我个人而言，会使用一个阵列或的ArrayList（根据您的需求）来存储int类型，然后使用while循环检查和接受的数字。 while（array.length < 3）do {int checking}。这样，如果数组中已经有3个值，那么脚本将不会运行，并且会一直运行，直到您获得3个值。