Connect to a password protected URL from Java program

By: Mike Cooper  

It is quite common for a Java program to connect to an URL and process its data or post data to it. Some URLs are password protected and can be connected to it only with a username and password. While manually accessing such sites, the browser will popup a login screen automatically for the user to key in username and password. But if you are connecting from a Java program then how can you do this login programmatically.

This sample Java program connects to password protected URL. The three arguments for this application is the password protected url to connect to, the username and the password. Take note that the System properties to set the proxy has to be changed to suit your own network proxy settings. If your network doesn't need a proxy then just set the proxy to false.

import java.util.*;
public class postto{

private void fetchURL (String urlString,String user,String pass) {

try {
URL url;
URLConnection urlConn;
DataOutputStream printout;
DataInputStream input;

Properties sysProperties = System.getProperties();

sysProperties.put("proxyHost", "");
sysProperties.put("proxyPort", "8080");
sysProperties.put("proxySet", "true");
url = new URL (urlString);
urlConn = url.openConnection();
urlConn.setDoInput (true);

urlConn.setDoOutput (true);

urlConn.setUseCaches (false);

urlConn.setRequestProperty ("Content-Type", "application/x-www-form-urlencoded");

printout = new DataOutputStream (urlConn.getOutputStream ());

String content = "USERNAME=" + URLEncoder.encode (user) + "&PASSWORD=" + URLEncoder.encode (pass);

printout.writeBytes (content);
printout.flush ();
printout.close ();

input = new DataInputStream (urlConn.getInputStream ());
FileOutputStream fos=new FileOutputStream("postto.txt");
String str;
while (null != ((str = input.readLine())))

if (str.length() >0)
fos.write(new String("\n").getBytes());
input.close ();
catch(MalformedURLException mue){ System.out.println (mue);}
catch(IOException ioe){ System.out.println (ioe);}
public static void main (String args[]) {
postto au=new postto();

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View Tutorial          By: mobile marketing at 2017-04-01 01:36:13

10. Hey thanks for your tutorial. But i'm getting an error when i run it.

Exception in th

View Tutorial          By: acc at 2013-07-25 14:30:50

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But if i had a Site with a security answer, how to impl

View Tutorial          By: Matze at 2013-07-19 16:08:46

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View Tutorial          By: Bharath at 2012-12-18 18:32:33

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View Tutorial          By: mohd at 2011-08-07 15:39:18

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View Tutorial          By: kumar at 2010-10-06 10:42:55

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View Tutorial          By: mahesh at 2009-08-26 01:31:40

16. Thanks for the help, however I am having some difficulty implementing it.
I keep getting the

View Tutorial          By: Andrew at 2008-09-30 14:51:14

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