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HttpUrlConnection addRequestProperty Method Not Passing Parameters

I have some working java code which does the following:

URL myUrl = new URL("http://localhost:8080/webservice?user=" + username + "&password=" + password + "&request=x");

HttpURLConnection myConnection = (HttpURLConnection) myUrl.openConnection();
myConnection.setRequestMethod("POST");

// code continues to read the response stream

However, I noticed that my webserver access log contained the plaintext password for all of the users who connected. I would like to get this out of the access log, but the webserver admins claim that this needs to be changed in my code and not via webserver config.

I tried changing the code to the following:

URL myUrl = new URL("http://localhost:8080/webservice");

HttpURLConnection myConnection = (HttpURLConnection) myUrl.openConnection();
myConnection.setRequestMethod("POST");
// start of new code
myConnection.setDoOutput(true);
myConnection.addRequestProperty("username", username);
myConnection.addRequestProperty("password", password);
myConnection.addRequestProperty("request", "x");

// code continues to read the response stream

Now the access log does not contain the username/password/request method. However, the webservice now throws an exception indicating that it didn't receive any username/password.

What did I do wrong in my client code? I also tried using "setRequestProperty" instead of "addRequestProperty" and it had the same broken behavior.

like image 492
David Avatar asked Mar 21 '13 15:03

David


1 Answers

I actually found the answer in another question on stackoverflow.

The correct code should be:

URL myUrl = new URL("http://localhost:8080/webservice");

HttpURLConnection myConnection = (HttpURLConnection) myUrl.openConnection();
myConnection.setRequestMethod("POST");
myConnection.setDoOutput(true);

DataOutputStream wr = new DataOutputStream(myConnection.getOutputStream ());
wr.writeBytes("username=" + username + "&password="+password + "&request=x");

// code continues to read the response stream
like image 55
David Avatar answered Nov 15 '22 10:11

David