How can I send POST data through url.openStream()?

Question

i'm looking for tutorial or quick example, how i can send POST data throw openStream.

My code is:

URL url = new URL("http://localhost:8080/test");
            InputStream response = url.openStream();
            BufferedReader reader = new BufferedReader(new InputStreamReader(response, "UTF-8"));

Could you help me ?

possible duplicate of [Sending HTTP POST Request In Java](http://stackoverflow.com/questions/3324717/sending-http-post-request-in-java) — Brian Roach, Jan 10 '12 at 14:10

score 8 · Answer 1 · edited Sep 16 '13 at 11:20

    URL url = new URL(urlSpec);
    HttpURLConnection connection = (HttpURLConnection) url.openConnection();
    connection.setRequestMethod(method);
    connection.setDoOutput(true);
    connection.setDoInput(true);

    // important: get output stream before input stream
    OutputStream out = connection.getOutputStream();
    out.write(content);
    out.close();        

            // now you can get input stream and read.
    BufferedReader reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));
    String line = null;

    while ((line = reader.readLine()) != null) {
        writer.println(line);
    }

+1 for answering the question instead of jumping to an alternative library. — Dev, Jan 10 '12 at 14:41

score 1 · Answer 2 · answered Jan 10 '12 at 14:06

Use Apache HTTP Compoennts http://hc.apache.org/httpcomponents-client-ga/

tutorial: http://hc.apache.org/httpcomponents-client-ga/tutorial/html/fundamentals.html

Look for HttpPost - there are some examples of sending dynamic data, text, files and form data.

score 0 · Answer 3 · answered Jan 10 '12 at 14:10

0

Apache HTTP Components in particular, the Client would be the best way to go. It absracts a lot of that nasty coding you would normally have to do by hand

answered Jan 10 '12 at 14:10

jere

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How can I send POST data through url.openStream()?

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