如何修复使用 "parseInt()" 的 if 语句，以便激活 else 部分并且我的程序不会失败？

Question

我写了一个程序，要求用户输入（数字），然后在确保两个值都是数字后告诉用户他们输入的两个数字中哪个更大。这是循环的。问题是，当用户输入无效的内容时，程序会关闭而不是引用我的 else 语句。我认为这是因为无法使用 parseInt() 处理字符串中的文本值，所以程序没有意识到文本值不是有效数字，而是失败了。如果可以帮助任何人解决问题，我正在使用 BlueJ 来制作这个程序。我也欢迎人们告诉我如何使我的程序更 efficient/easier 编码（我是初学者）。

import java.util.Scanner;
public class TemperatureDriver {
    public static void main(String[] args) {
        while (true) {
            Scanner keyInput = new Scanner(System.in); 
            System.out.print("Enter a number"+ "\n");
            String number_one = keyInput.next();

            if (Integer.parseInt(number_one) <=0 || Integer.parseInt(number_one) > 0) {
                System.out.print("Enter another number" + "\n");
                String number_two = keyInput.next();     

                if(Integer.parseInt(number_two) <=0 || Integer.parseInt(number_two) > 0){

                    if (Integer.parseInt(number_one) > Integer.parseInt(number_two)) {
                        System.out.print(number_one + " is greater than " + number_two + "\n" + "\n");
                    } else if(Integer.parseInt(number_one) < Integer.parseInt(number_two)){
                        System.out.print(number_one + " is less than " + number_two + "\n" + "\n");
                    } else if(Integer.parseInt(number_one) == Integer.parseInt(number_two)){
                        System.out.print(number_one + " is equal to " + number_two + "\n" + "\n");
                    }                

                } else {
                    System.out.println("Invalid number!"+ "\n");
                }  

            } else {
                System.out.println("Invalid number!"+ "\n");
            }  

        }
    }
}

Answer 1

您调用 parseInt 的代码应嵌入到 try 块中，该块后应跟一个 catch 块以捕获 NumberFormatException。

try {
    if (Integer.parseInt(number_one) <= 0 || Integer.parseInt(number_one) > 0) {
      System.out.print("Enter another number" + "\n");
      String number_two = keyInput.next();

      if (Integer.parseInt(number_two) <= 0 || Integer.parseInt(number_two) > 0) {

        if (Integer.parseInt(number_one) > Integer.parseInt(number_two)) {
          System.out.print(number_one + " is greater than " + number_two + "\n" + "\n");
        } else if (Integer.parseInt(number_one) < Integer.parseInt(number_two)) {
          System.out.print(number_one + " is less than " + number_two + "\n" + "\n");
        } else if (Integer.parseInt(number_one) == Integer.parseInt(number_two)) {
          System.out.print(number_one + " is equal to " + number_two + "\n" + "\n");
        }

      } else {
        System.out.println("Invalid number!" + "\n");
      }

    } else {
      System.out.println("Invalid number!" + "\n");
    }
  } catch (NumberFormatException e) {
    System.out.println("Not a valid number");
  }

如何修复使用 "parseInt()" 的 if 语句，以便激活 else 部分并且我的程序不会失败？

How do I fix an if statement that uses "parseInt()" so that the else portion is activated and my program doesn't fail?

java

bluej