Generate a HTML POST url and open the page.

Question

Hello,
I need to generate a HTML POST url with 4 parameters.
I tried like this.
String URLtobeOpened = "<url>?param1=val1&param2=val2&param3=val3&param4=val4";
          HttpServletResponse resp = request.getServletResponse(true);
          try {
               resp. sendRedirect(URLtobeOpened);
          } catch (IOException e1) {
               // TODO Auto-generated catch block
               e1.printStackTrace();
This is showing me a login prompt, instead of directly logging me in.
Also when I checked using HTTPWatch, the request is showing as a GET, but I need it as a POST.
How to code it in the right way.
Thanks
Srinivas

Answer 1

Here is what I tried sofar
I am using this jar commons-httpclient-3.1.jar
import org.apache.commons.httpclient.HttpClient;
import org.apache.commons.httpclient.methods.PostMethod;
          String url = "https://www.independentinc.com/scripts/insupport.exe/db_connect";
            try {
             HttpClient client = new HttpClient();
             PostMethod method = new PostMethod( url );
               // Configure the form parameters
               method.addParameter( "param1", "val1" );
               method.addParameter( "param2", "val2" );
               method.addParameter( "param3", "val3" );
               method.addParameter( "param4", "val4" );
               // Execute the POST method
             int statusCode = client.executeMethod( method );
             if( statusCode != -1 ) {
                String contents = method.getResponseBodyAsString();
                method.releaseConnection();
//                System.out.println( contents );
            catch( Exception e ) {
             e.printStackTrace();
            }I am getting a weird exception
java.lang.NoClassDefFoundError: org/apache/commons/logging/LogFactory at
HttpClient client = new HttpClient();How to fix this exception...
Is this the right approach at all...