• Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 366
  • Last Modified:

Sending a xml file/String to remote servlet.

Hi All
I am running a query and getting the resultset and using that i am formulating a string in xml format.My query here is i need to send that string/xmlfile to a remote servlet.
help me out how to send that to remote servlet(I need to send in post method only)
0
krishnamukuntha
Asked:
krishnamukuntha
  • 4
  • 3
1 Solution
 
KuldeepchaturvediCommented:
http://www.experts-exchange.com/Web/Web_Languages/JSP/Q_20665134.html

Only thing that you might want to change in the code is the method that I am setting to GET... you will have to convert it to POST
0
 
krishnamukunthaAuthor Commented:
I want to send it from my servlet to remote
i hope your code is doing on other way...
0
 
KuldeepchaturvediCommented:
nopes its sending to the remote URL and reading the response back
0
Free Tool: Site Down Detector

Helpful to verify reports of your own downtime, or to double check a downed website you are trying to access.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

 
krishnamukunthaAuthor Commented:
let me rephrase my query:
On Click of a button i am hittin the DB and creating report in xml format by looping the Resultset.
Now i ned to send the created xml file to repmote servlet.
I also need to know how the remort server can get my file in its doPost Method
0
 
KuldeepchaturvediCommented:
BufferedReader in=null;
                        
//                  String clientUrl = POINTProperties.getProperty("CFWURL");
                  String clientUrl = CFWURL;
                  clientUrl +="&setSystemProperties=AS400ConnectUserIdTREE("+usr+"),AS400ConnectUserPasswordTREE("+pwd+")";
//                  clientUrl +="&setOverrideProperties=AS400ConnectUserId("+usr+"),AS400ConnectUserPassword("+pwd+")";
                  URL cfwUrl = new URL(clientUrl);
                  
                  java.net.HttpURLConnection con = null;
                  //Open a connection to the specified url
                  try
                  {  
                        con =(java.net.HttpURLConnection) cfwUrl.openConnection();
                        con.setRequestProperty("Content-Type","text/xml");
                        con.setDoOutput(true);
                        con.setDoInput(true);
                        con.setRequestMethod("POST");
                        //Post the XML transaction to the system.
                        con.getOutputStream().write(xml.getBytes());//this code will post your XML to
whatever url you have in cfwURL...
                        in = new BufferedReader(new InputStreamReader(con.getInputStream()));
                  }
                  //Catch any errors that may happen is this process.
                  catch(ProtocolException pe)
                  {
                        System.out.println("Received a Protocol Exception in doRequest"+pe);
                        return null;
                  }
                  
                  catch (FileNotFoundException exception)
                  {
                        System.out.println("Exception in PM-doRequest"+exception);
                        InputStream err = con.getErrorStream();
                        in = new BufferedReader(new InputStreamReader(err));
                  }
                  catch(IOException ioe)
                  {
                        System.out.println("Received an Exception in PM-doRequest"+ioe);
                        ioe.printStackTrace();
                        return null;
                  
                  }
            //Try reading the response from the system. catch any errors that may happen in the process.      
                  try
                  {
                  StringBuffer response = new StringBuffer();
                  String line;
      
                  while ((line = in.readLine()) != null)
                        response.append(line);
                  in.close();
                  return response.toString();
                  }
                  catch(IOException ioex)
                  {
                  System.out.println("ioex in PM-doRequest"+ioex);
                  ioex.printStackTrace();
                  return null;
                  }
                  catch(Exception e1)
                  {
                        System.out.println("Error in PM-doRequest"+e1);
                        e1.printStackTrace();
                        return null;
                  }
            }
            catch(Exception e2)
            {
                  return null;
            }
      }


in the remote servlet all you have to do in doPost is
request.getInputStream(); //this stream will have the XML that you have posted
0
 
krishnamukunthaAuthor Commented:
thanks kuldeep
whole nation will proud of you :)
0
 
KuldeepchaturvediCommented:
:-) glad to be of help
thanks for the 'A' grade...
0

Featured Post

Hire Technology Freelancers with Gigs

Work with freelancers specializing in everything from database administration to programming, who have proven themselves as experts in their field. Hire the best, collaborate easily, pay securely, and get projects done right.

  • 4
  • 3
Tackle projects and never again get stuck behind a technical roadblock.
Join Now