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Solved

Sending a xml file/String to remote servlet.

Posted on 2004-04-14
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331 Views
Last Modified: 2010-04-01
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)
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Question by:krishnamukuntha
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7 Comments
 
LVL 19

Expert Comment

by:Kuldeepchaturvedi
ID: 10823001
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
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LVL 2

Author Comment

by:krishnamukuntha
ID: 10823177
I want to send it from my servlet to remote
i hope your code is doing on other way...
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LVL 19

Expert Comment

by:Kuldeepchaturvedi
ID: 10823185
nopes its sending to the remote URL and reading the response back
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LVL 2

Author Comment

by:krishnamukuntha
ID: 10823224
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
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LVL 19

Accepted Solution

by:
Kuldeepchaturvedi earned 250 total points
ID: 10823509
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
 
LVL 2

Author Comment

by:krishnamukuntha
ID: 10823856
thanks kuldeep
whole nation will proud of you :)
0
 
LVL 19

Expert Comment

by:Kuldeepchaturvedi
ID: 10823869
:-) glad to be of help
thanks for the 'A' grade...
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