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Home/ Questions/Q 4252580
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Editorial Team
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Editorial Team
Asked: May 21, 20262026-05-21T04:48:51+00:00 2026-05-21T04:48:51+00:00

I have two web services running on my tomcat lets call them X and

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I have two web services running on my tomcat lets call them X and Y.

when Y is requested by X in the end Y have a String with a XML inside and now i want to return this string of XML to X.

How can i do this ?

is there anyway to make a special request that X servlet waits for a response ? or do i really need to redirect ?

thx for your time.

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  1. Editorial Team
    Editorial Team
    2026-05-21T04:48:52+00:00Added an answer on May 21, 2026 at 4:48 am

    Since you have a java tag, I will assume you are using java servlets.

    X gets a XML, that should go to y?
    Well the easiest way is to for X to create a request to y and forward the response back to the person who made the request.

    You can use the HTTPClient class for that.

    so once you send the request from X to Y, X will wait until Y has responded.

    If your unsure how to do this checkout the java docs http://download.oracle.com/javase/1.5.0/docs/api/java/net/HttpURLConnection.html

    Make sure u check variable sets and parameters, i pulled this from some of my code

    Code:

    try {
            URL url = new URL(server);
    
            HttpURLConnection con;
            con=(HttpURLConnection) url.openConnection();
            con.setRequestProperty("Content-type", "text/xml; charset=UTF-8");
            con.setRequestMethod("POST");
            con.setDoOutput(true);
            con.setDoInput(true);
    
    
            OutputStream out = con.getOutputStream();
            Writer writer = new OutputStreamWriter(out, "UTF-8");
    
            writer.write(xml);
    
    
    
            writer.flush();
            writer.close();
    
            InputStream is= con.getInputStream();
    
    //This gets sent to the client
                return set_courses(is);
    
    
        } catch (Exception e){
            e.printStackTrace();
            status_message= "Custom 1: "+e.getMessage();
            return false;
        }
    
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