如果发生用户输入错误，如何防止Java代码处理？

Question

我正在编写使用GUI的温度转换器。我将其设置为如果用户输入字符串或无效的温度，则在输入框下方显示错误标签。但是，该代码仍将像用户输入0一样进行处理。它将仍然输出转换。如何在下面的课程中阻止它这样做？

package gui;
import javax.swing.*;
import java.awt.*;

public class JDataInput
    extends JPanel
{
    private JLabel lblInput;
    private JTextField txtInput;
    private JLabel lblMessage;

    public JDataInput()
    {

    }

    public JDataInput(String caption)
    {

        lblInput = new JLabel(caption);
        txtInput = new JTextField(20);
        lblMessage = new JLabel("");

        JPanel row1 = new JPanel();
        JPanel row2 = new JPanel();

        row1.add(lblInput);
        row1.add(txtInput);
        row2.add(lblMessage);

        this.add(row1);
        this.add(row2);

    }

    public double getDoubleValue()
    {
        double returnValue = 0;
        lblMessage.setText(" ");

        try
        {
            returnValue =   Double.parseDouble(txtInput.getText());

        }
        catch (NumberFormatException nfex)
        {
            lblMessage.setText(nfex.toString());
            lblMessage.setForeground(Color.RED);
        }
        return returnValue;
    }

    public double getIntValue()
    {
        int returnValue = 0;
        lblMessage.setText(" ");
        try
        {
            returnValue =   Integer.parseInt(txtInput.getText());
        }
        catch (NumberFormatException nfex)
        {
            lblMessage.setForeground(Color.RED);
            lblMessage.setText(nfex.toString());
        }
        return returnValue;
    }

}

Answer 1

从catch块中返回空值，例如

 public double getDoubleValue()
    {
        double returnValue = 0;
        lblMessage.setText(" ");

        try
        {
            returnValue =   Double.parseDouble(txtInput.getText());

        }
        catch (NumberFormatException nfex)
        {
            lblMessage.setText(nfex.toString());
            lblMessage.setForeground(Color.RED);
            return null;
        }
        return returnValue;
    }

这将从函数中退出。