Java 异常阻止关闭程序（单击 X 按钮时）

Question

我使用 JOptionPane 构建了一个随机数生成器。我编写的异常阻止用户在单击 X 按钮时退出程序。我做了很多研究并尝试了几件事，但似乎没有任何效果。

这是我的代码：

import javax.swing.JOptionPane;
import java.util.Random;

public class Generate {
    private int number;
    private int min;
    private int max;
    private int repeat;
    Random no = new Random();
    int x = 1;

    void generateNumber() {
        do {
            try {
                String from = (String) JOptionPane.showInputDialog(null, "Welcome to Random Number Generator!\n\nPlease insert your number range.\n\nFrom:", "Random Number Generator", JOptionPane.QUESTION_MESSAGE, null, null, "Enter Number");
                min = Integer.parseInt(from);

                String to = (String) JOptionPane.showInputDialog(null, "To:", "Random Number Generator", JOptionPane.QUESTION_MESSAGE, null, null, "Enter Number");
                max = Integer.parseInt(to);
                System.out.println();

                String count = (String) JOptionPane.showInputDialog(null, "How many numbers would you like?", "Random Number Generator", JOptionPane.QUESTION_MESSAGE, null, null, "Enter Number");
                repeat = Integer.parseInt(count);
                System.out.println();

                for (int counter = 1; counter <= repeat; counter++) {
                    number = no.nextInt(max - min + 1) + min;
                    JOptionPane.showMessageDialog(null, "Random number #" + counter + ": " + number, "Random Number Generator", JOptionPane.PLAIN_MESSAGE);
                }
               x = 2;
            } catch (NumberFormatException e) {
                  JOptionPane.showMessageDialog(null, "INPUT ERROR: please insert a number", "Random Number Generator", JOptionPane.ERROR_MESSAGE);
            } catch (Exception e) {
                  JOptionPane.showMessageDialog(null, "INPUT ERROR: the second number needs to be higher than the first", "Random Number Generator", JOptionPane.ERROR_MESSAGE);
                }
            } while(x == 1);
        }
}

主要：

  class RandomNumber {
    public static void main(String[] args) {
        Generate obj = new Generate();
        obj.generateNumber();
    }
}

This is what happens when I try to close the program

Answer 1

您没有在 showInputDialog() 次调用后测试 from 值。

例如这里：

String from = (String) JOptionPane.showInputDialog(null,...

此调用后，您直接与

链接

min = Integer.parseInt(from);

无论 from 的值如何。
如果 from 是 null 你完成这个 catch 因为 null 不是一个数字 :

  catch (NumberFormatException e) {
        JOptionPane.showMessageDialog(null, "INPUT ERROR: please insert a number", "Random Number Generator",
        JOptionPane.ERROR_MESSAGE);
    }

并且 x 的值仍然是 1。所以循环条件还是true.

为了解决您的问题，您只需测试showMessageDialog()返回的值，如果值为null，让用户退出该方法。

每次检索用户输入并希望允许用户退出时添加此代码：

if (from == null) {
    return;
}

Java 异常阻止关闭程序（单击 X 按钮时）

Java Exceptions Prevent Closing The Program (when clicking X button)

java

exception

try-catch

throw

joptionpane