如何在Java中捕获异常？ [英] How to catch an exception in Java?

问题描述

如何在Java中捕获异常？我有一个程序可以接受整数值的用户输入。现在如果用户输入无效值，它会抛出一个 java.lang.NumberFormatException 。如何捕获该异常？

How to catch an exception in Java? I have a program that accepts user input which is of integer value. Now if the user enters an invalid value, it throws a java.lang.NumberFormatException. How do I catch that exception?

    public void actionPerformed(ActionEvent e) {
        String str;
        int no;
        if (e.getSource() == bb) {
            str = JOptionPane.showInputDialog("Enter quantity");
            no = Integer.parseInt(str);
 ...

推荐答案

try {
   int userValue = Integer.parseInt(aString);
} catch (NumberFormatException e) {
   //there you go
}

，特别是在您的代码中：

and specifically in your code:

public void actionPerformed(ActionEvent e) {
    String str;
    int no;
    //------------------------------------
    try {
       //lots of ifs here
    } catch (NumberFormatException e) {
        //do something with the exception you caught
    }

    if (e.getSource() == finish) {
        if (message.getText().equals("")) {
            JOptionPane.showMessageDialog(null, "Please Enter the Input First");
        } else {
            leftButtons();

        }
    }
    //rest of your code
}

这篇关于如何在Java中捕获异常？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！