Question

I am a newbie in java and was trying android development . The following code generated malformedURLException.can someone help me identify the exception.
Any hint would be highly helpful

package com.example.helloandroid;

import android.app.Activity;
//import android.widget.TextView;
import android.os.Bundle;
import java.net.*;
import java.io.*;
import android.widget.TextView;

public class HelloAndroid extends Activity {
    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        TextView tv = new TextView(this);
        String outdata = "";
        URL url_g = new URL("http://www.google.com/");
        URLConnection ukr = url_g.openConnection();

        BufferedReader in = new BufferedReader(new InputStreamReader(ukr.getInputStream()));
        String inputLine;
        while ((inputLine = in.readLine()) != null)
            outdata += inputLine;
        in.close();
       tv.setText(outdata);
       setContentView(tv);
    }
}

Answer 1

This is because URL constructor ( new URL(“http://www.google.com/”) ) throws a Checked Exception, in this case MalformedURLException, which means that you have to catch it or declare it in your method.

Use a try/catch or the throws clause in the onCreate method.

try/catch solution:

public void onCreate(Bundle savedInstanceState) {
    ...
    try {
        URL url_g = new URL("http://www.google.com/");
    } catch(MalformedURLException e) {
        //Do something with the exception.
    }
    ...
}

Throws solution:

@Override
public void onCreate(Bundle savedInstanceState) throws MalformedURLException {
    ...
    URL url_g = new URL("http://www.google.com/");
    ...
}

Check this for more information on exceptions.

http://download.oracle.com/javase/tutorial/essential/exceptions/