Alex Rivera | Logout

Why do I get "Exception; must be caught or declared to be thrown" when I try to compile my Java code?

Asked 2009-05-26T02:22:57.050
51

Consider:

import java.awt.*;

import javax.swing.*;
import java.awt.event.*;
import javax.crypto.*;
import javax.crypto.spec.*;
import java.security.*;
import java.io.*;


public class EncryptURL extends JApplet implements ActionListener {

    Container content;
    JTextField userName = new JTextField();
    JTextField firstName = new JTextField();
    JTextField lastName = new JTextField();
    JTextField email = new JTextField();
    JTextField phone = new JTextField();
    JTextField heartbeatID = new JTextField();
    JTextField regionCode = new JTextField();
    JTextField retRegionCode = new JTextField();
    JTextField encryptedTextField = new JTextField();

    JPanel finishPanel = new JPanel();


    public void init() {

        //setTitle("Book - E Project");
        setSize(800, 600);
        content = getContentPane();
        content.setBackground(Color.yellow);
        content.setLayout(new BoxLayout(content, BoxLayout.Y_AXIS));

        JButton submit = new JButton("Submit");

        content.add(new JLabel("User Name"));
        content.add(userName);

        content.add(new JLabel("First Name"));
        content.add(firstName);

        content.add(new JLabel("Last Name"));
        content.add(lastName);

        content.add(new JLabel("Email"));
        content.add(email);

        content.add(new JLabel("Phone"));
        content.add(phone);

        content.add(new JLabel("HeartBeatID"));
        content.add(heartbeatID);

        content.add(new JLabel("Region Code"));
        content.add(regionCode);

        content.add(new JLabel("RetRegionCode"));
        content.add(retRegionCode);

        content.add(submit);

        submit.addActionListener(this);
    }


    public void actionPerformed(ActionEvent e) {

        if (e.getActionCommand() == "Submit"){

            String subUserName = userName.getText();
            String subFName = firstName.getText();
            String subLName = lastName.getText();
java exception
Edit
Report

1 Answer

1

You'll need to decide how you'd like to handle exceptions thrown by the encrypt method.

Currently, encrypt is declared with throws Exception - however, in the body of the method, exceptions are caught in a try/catch block. I recommend you either:

  • remove the throws Exception clause from encrypt and handle exceptions internally (consider writing a log message at the very least); or,
  • remove the try/catch block from the body of encrypt, and surround the call to encrypt with a try/catch instead (i.e. in actionPerformed).

Regarding the compilation error you refer to: if an exception was thrown in the try block of encrypt, nothing gets returned after the catch block finishes. You could address this by initially declaring the return value as null:

public static byte[] encrypt(String toEncrypt) throws Exception{
  byte[] encrypted = null;
  try {
    // ...
    encrypted = ...
  }
  catch(Exception e){
    // ...
  }
  return encrypted;
}

However, if you can correct the bigger issue (the exception-handling strategy), this problem will take care of itself - particularly if you choose the second option I've suggested.

answered 2009-05-26T02:43:38.977

Your Answer