java - 如果用户输入随机字母，如何给出对话框错误 JOptionPane

Question

所以我是java新手，我正在尝试使用try和catch。例如，如果我询问用户有多少葡萄，并且他们输入了一堆字母，它将显示一个错误对话框，而不仅仅是出现系统错误。我可以使用扫描仪来完成此操作，但不能使用 JOptionPane。我真的希望出现一个对话框，这就是我尝试使用 JOptionPane.showInputDialog 的原因。

扫描仪可以工作...=

import java.util.Scanner;

class test {
    public static void main (String[] args)
    {

        Scanner input = new Scanner(System.in);

        System.out.println("How many grapes do u have?");
        int grapes = 1;
        try
        {
            grapes = input.nextInt();
        }
        catch (Exception e)
        {
            System.out.println("Good job Sherlock you broke the program");
            return;
        }
        int mg;

        if (grapes >= 100)
            mg = 1;

        else
            mg = 2;

        switch (mg){
            case 1: System.out.println("You got a lot of grapes");
                break;
            case 2: System.out.println("You brarely got any grapes");
                break;
        }
    }
}

JOptionPane 不起作用...

import javax.swing.JOptionPane;

public class bday
{
    public static void main(String[] args)
    {
        String age = "0";
        try
        {
            age = JOptionPane.showInputDialog("What was your age yesterday?");
        }
        catch(Exception e)
        {
            JOptionPane.showMessageDialog(null, "Thanks a lot, you broke it. CYA later.");
            return;
        }
        int iage = Integer.parseInt(age);

        String bday = "0";

         try
         {
             bday = JOptionPane.showInputDialog("Was yesterday your B-Day? (True or False)");
         }
         catch (Exception e)
         {
            JOptionPane.showMessageDialog(null, "WHY U MESS UP PROGRAM???.... BYE BYE!!");
            return;
         }
         boolean bage = Boolean.parseBoolean(bday);
        if (bage == true){
            iage += 1;
            JOptionPane.showMessageDialog(null, "You are now " + iage);
        }
        else if (bage == false){
            JOptionPane.showMessageDialog(null, "Happy unbirthday!");
        }
        if (iage ==10){
            JOptionPane.showMessageDialog(null, "Congrats, double digits!");
        }
        if (iage >19){
            JOptionPane.showMessageDialog(null,  "U aint a Teenager");
        }
        else if (iage < 13)
            JOptionPane.showMessageDialog(null,  "U aint a Teenager");          

    }
}

Answer 1

你正在尝试/捕获错误的东西。您应该将解析语句放在 try block 中，因为这会引发异常。

例如，不是

    String age = "0";
    try
    {
        age = JOptionPane.showInputDialog("What was your age yesterday?");
    }
    catch(Exception e)
    {
        JOptionPane.showMessageDialog(null, "Thanks a lot, you broke it. CYA later.");
        return;
    }
    int iage = Integer.parseInt(age);

而是:

String age = JOptionPane.showInputDialog("What was your age yesterday?");
try {
  iage = Integer.parseInt(age);
} catch (NumberFormatException nfe) {
  // show error
}  

此外，您应该避免捕获Exception，而应该只捕获特定的异常，此处为NumberFormatException。

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编辑
在评论中你问:

One more question, should i do the same with the boolean?

在解析时， boolean 类型有点棘手(在我看来)。要了解 Boolean.parseBoolean(...) 的工作原理，请查看 Boolean API ，特别是 parseBoolean 方法。您将看到，如果输入的文本没有意义，它不会抛出 NumberFormatException。 API 会告诉您实际上返回了什么。 try/catch block 在这里不起作用。如果您需要捕获错误，请考虑使用 String 的 equalsIgnoreCase(...)。