Http client Post xml file in java

Question

I need to send a xml file to the following link\

    http://14.140.66.142:80/MSMQ/private$/votes

This is my code.

   URL url = new URL("http://14.140.66.142:80/MSMQ/private$/votes");
    URLConnection con = url.openConnection();
    String document = "C:\\Documents and Settings\\Nagra\\My Documents\\Responseserver\\workingVoting\\VoteSubmitter\\Body.xml";

    FileReader fr = new FileReader(document);
    // specify that we will send output and accept input
    con.setDoInput(true);
    con.setDoOutput(true);
    char[] buffer = new char[1024*10];

    int b_read = 0;

    if ((b_read = fr.read(buffer)) != -1)

    {
        con.setRequestHeader ( "Content-Type", "text/xml" );
        con.setRequestProperty("SOAPAction","MSMQMessage");
        con.setRequestProperty("Proxy-Accept","NonInteractiveClient" );
        con.setRequestProperty("CONNECTION", "close");
        con.setRequestProperty("CACHE-CONTROL", "no-cache");
        con.setRequestProperty("USER-AGENT", "OpenTV-iAdsResponder_1_0");
        OutputStreamWriter writer = new OutputStreamWriter( con.getOutputStream() );
        writer.write(buffer, 0, b_read);
        PrintWriter pw = new PrintWriter(con.getOutputStream());
        pw.write(buffer, 0, b_read);
       pw.close();
        System.out.println("written");


  }
  catch( Throwable t )
{
    t.printStackTrace( System.out );
}

  }
  }

I don't Know whether it is right code.If i run this code I am not able to receive the xml file on the server side.Can anyone help me where i gone wrong in my code.

score 1 · Answer 1 · answered Oct 13 '11 at 09:42

1

Below is a sample POST operation:

URL url = new URL("http://14.140.66.142:80/MSMQ/private$/votes");
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setDoOutput(true);
connection.setInstanceFollowRedirects(false);
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type", "application/xml");

OutputStream os = connection.getOutputStream();

TransformerFactory tf = TransformerFactory.newInstance();
Transformer transformer = tf.newTransformer();
FileReader fileReader = new FileReader("C:\\Documents and Settings\\Nagra\\My Documents\\Responseserver\\workingVoting\\VoteSubmitter\\Body.xml");
StreamSource source = new StreamSource(fileReader);
StreamResult result = new StreamResult(os);
transformer.transform(source, result);

os.flush();
connection.getResponseCode();
connection.disconnect();

answered Oct 13 '11 at 09:42

bdoughan

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This is the error which i Got while running your code. org.xml.sax.SAXParseException: Content is not allowed in prolog. at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse(AbstractSAXParser.java:1231) at com.sun.org.apache.xalan.internal.xsltc.trax.TransformerImpl.transformIdentity(TransformerImpl.java:609) at com.sun.org.apache.xalan.internal.xsltc.trax.TransformerImpl.transform(TransformerImpl.java:707) at com.sun.org.apache.xalan.internal.xsltc.trax.TransformerImpl.transform(TransformerImpl.java:313) at hhtprequest.main(hhtprequest.java:52) – bharathi Oct 13 '11 at 09:59
@bharathi - Can you try the transform outside the HTTP operation (set the output to System.out)? This code has been successful for others:http://stackoverflow.com/questions/6665769/jersey-rest-client-how-to-add-xml-file-to-the-body-of-post-request/6667391#6667391 – bdoughan Oct 13 '11 at 10:08

score 0 · Answer 2 · answered Sep 25 '14 at 14:48

There are a couple of issues with the code you have posted. First, you are reading only 1024*10 characters and you are not sending the whole file if the file has more characters. Second, you are writing the content more than once. Change the code something similar to this.

URL url = new URL("http://14.140.66.142:80/MSMQ/private$/votes");
HttpURLConnection con = (HttpURLConnection)url.openConnection();
String document = "C:\\Documents and Settings\\Nagra\\My Documents\\Responseserver\\workingVoting\\VoteSubmitter\\Body.xml";

FileReader fr = new FileReader(document);
// specify that we will send output and accept input
con.setDoInput(true);
con.setDoOutput(true);
char[] buffer = new char[1024*10];
int b_read = 0;
con.setRequestProperty ( "Content-Type", "text/xml" );
con.setRequestProperty("SOAPAction","MSMQMessage");
con.setRequestProperty("Proxy-Accept","NonInteractiveClient" );
con.setRequestProperty("CONNECTION", "close");
con.setRequestProperty("CACHE-CONTROL", "no-cache");
con.setRequestProperty("USER-AGENT", "OpenTV-iAdsResponder_1_0");
OutputStreamWriter writer = new OutputStreamWriter( con.getOutputStream() );
while ((b_read = fr.read(buffer)) != -1) {
    writer.write(buffer, 0, b_read);
}
writer.flush();
writer.close();
fr.close();
int i = con.getResponseCode();
con.disconnect();
System.out.println(String.format("written with response code: %d",i));

Http client Post xml file in java

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