Java 如果有效整数不是'，我如何让程序以错误终止；你没进去吗？

对于学校，我正在制作一个程序，用户输入三个整数，程序找到这三个整数的乘积，并将结果输出给用户。老师要求我使用JOptionPane类。输入无效整数时，如何使程序以错误终止。还有，如何在java窗口中输出答案？提前谢谢

import javax.swing.JOptionPane;

public class ASTheProductofThreeGUI {

    public static void main(String[] args) {

        //initializes variable "answer" of type integer

        //prompts the user to enter their first integer for the product of three
        int value1 = Integer.parseInt(JOptionPane.showInputDialog("Enter your first"
                + " value as an integer")); 

        //prompts the user to enter their second integer
        int value2 = Integer.parseInt(JOptionPane.showInputDialog("Enter your second"
                + " value as an integer")); 

        //prompts the user to enter their third integer
        int value3 = Integer.parseInt(JOptionPane.showInputDialog("Enter your third"
                + " value as an integer")); 

        int answer = value1 * value2 * value3; 

---

如果字符串不是有效的整数，Integer.parseInt将引发异常。您应该处理NumberFormatException，然后调用System.exit

import javax.swing.JOptionPane;

public class ASTheProductofThreeGUI {

public static void main(String[] args) {

    //initializes variable "answer" of type integer

    //prompts the user to enter their first integer for the product of three
try {
    int value1 = Integer.parseInt(JOptionPane.showInputDialog("Enter your first"
            + " value as an integer")); 

    //prompts the user to enter their second integer
    int value2 = Integer.parseInt(JOptionPane.showInputDialog("Enter your second"
            + " value as an integer")); 

    //prompts the user to enter their third integer
    int value3 = Integer.parseInt(JOptionPane.showInputDialog("Enter your third"
            + " value as an integer")); 
 catch (NumberFormatException e){
    e.printStackTrace("Invalid Integer entered.");
    System.exit(0);
 }
    int answer = value1 * value2 * value3; 

---

同样，作为问题询问家庭作业也是不受欢迎的，因此我将留下您的第二个问题让您去了解。

如果字符串不是有效的整数，Integer.parseInt将引发异常。您应该处理NumberFormatException，然后调用System.exit

import javax.swing.JOptionPane;

public class ASTheProductofThreeGUI {

public static void main(String[] args) {

    //initializes variable "answer" of type integer

    //prompts the user to enter their first integer for the product of three
try {
    int value1 = Integer.parseInt(JOptionPane.showInputDialog("Enter your first"
            + " value as an integer")); 

    //prompts the user to enter their second integer
    int value2 = Integer.parseInt(JOptionPane.showInputDialog("Enter your second"
            + " value as an integer")); 

    //prompts the user to enter their third integer
    int value3 = Integer.parseInt(JOptionPane.showInputDialog("Enter your third"
            + " value as an integer")); 
 catch (NumberFormatException e){
    e.printStackTrace("Invalid Integer entered.");
    System.exit(0);
 }
    int answer = value1 * value2 * value3; 

---

同样，作为一个问题问家庭作业也是不被接受的，所以我将留下你的第二个问题让你去发现。

smokeyrobot，让我扩展你的建议，并给Samuel一个关于干净编码的提示。正如Robert C.Martin在他的优秀著作《干净的代码》中所说的那样，try/catch块是丑陋的，它混淆了错误处理与正常处理混合的代码结构。下面是我对塞缪尔问题的建议：

public class JOptionPaneExample {

public static void main(String[] args){
    try {
        productOfThreeNumbers();
    } catch (NumberFormatException e) {
        e.printStackTrace();
        showErrorDialog();
        System.exit(0);
    }

}

private static void productOfThreeNumbers() {
    //initializes variable "answer" of type integer

    //prompts the user to enter their first integer for the product of three
    int value1 = Integer.parseInt(JOptionPane.showInputDialog("Enter your first"
            + " value as an integer"));

    //prompts the user to enter their second integer
    int value2 = Integer.parseInt(JOptionPane.showInputDialog("Enter your second"
            + " value as an integer"));

    //prompts the user to enter their third integer
    int value3 = Integer.parseInt(JOptionPane.showInputDialog("Enter your third"
            + " value as an integer"));

    int answer = value1 * value2 * value3;
    JOptionPane.showMessageDialog(null, String.format("%d * %d * %d = %d", value1, value2, value3, answer));
}

private static void showErrorDialog() {
    //Put here code to show a friendly error message...
}}

---

smokeyrobot，让我扩展你的建议，并给Samuel一个关于干净编码的提示。正如Robert C.Martin在他的优秀著作《干净的代码》中所说的那样，try/catch块是丑陋的，它混淆了错误处理与正常处理混合的代码结构。下面是我对塞缪尔问题的建议：

public class JOptionPaneExample {

public static void main(String[] args){
    try {
        productOfThreeNumbers();
    } catch (NumberFormatException e) {
        e.printStackTrace();
        showErrorDialog();
        System.exit(0);
    }

}

private static void productOfThreeNumbers() {
    //initializes variable "answer" of type integer

    //prompts the user to enter their first integer for the product of three
    int value1 = Integer.parseInt(JOptionPane.showInputDialog("Enter your first"
            + " value as an integer"));

    //prompts the user to enter their second integer
    int value2 = Integer.parseInt(JOptionPane.showInputDialog("Enter your second"
            + " value as an integer"));

    //prompts the user to enter their third integer
    int value3 = Integer.parseInt(JOptionPane.showInputDialog("Enter your third"
            + " value as an integer"));

    int answer = value1 * value2 * value3;
    JOptionPane.showMessageDialog(null, String.format("%d * %d * %d = %d", value1, value2, value3, answer));
}

private static void showErrorDialog() {
    //Put here code to show a friendly error message...
}}