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Home/ Questions/Q 693235
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T02:43:58+00:00 2026-05-14T02:43:58+00:00

I would like to invoke a webservice via Android. I need to POST some

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I would like to invoke a webservice via Android. I need to POST some XML to a URL via HTTP.
I found this snipped for sending a POST, but i dont know how to include/add the XML data itself.

public void postData() {
         // Create a new HttpClient and Post Header  
         HttpClient httpclient = new DefaultHttpClient();  
         HttpPost httppost = new HttpPost("http://10.10.4.35:53011/");

         try {  
             // Add your data  
             List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);  
             nameValuePairs.add(new BasicNameValuePair("Content-Type", "application/soap+xml"));               
             httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs)); 
                 // Where/how to add the XML data?


             // Execute HTTP Post Request  
             HttpResponse response = httpclient.execute(httppost);  

         } catch (ClientProtocolException e) {  
             // TODO Auto-generated catch block  
         } catch (IOException e) {  
             // TODO Auto-generated catch block  
         }  
     }

This is the complete POST message that i need to imitate:

POST /a8103e90-f1e3-11dd-bfdb-8b1fcff1a110 HTTP/1.1
Host: 10.10.4.35:53011
Content-Type: application/soap+xml
Content-Length: 602

<?xml version='1.0' encoding='UTF-8' ?>
<s12:Envelope xmlns:s12="http://www.w3.org/2003/05/soap-envelope" xmlns:wsa="http://schemas.xmlsoap.org/ws/2004/08/addressing">
  <s12:Header>
    <wsa:MessageID>urn:uuid:fc061d40-3d63-11df-bfba-62764ccc0e48</wsa:MessageID>
    <wsa:Action>http://schemas.xmlsoap.org/ws/2004/09/transfer/Get</wsa:Action>
    <wsa:To>urn:uuid:a8103e90-f1e3-11dd-bfdb-8b1fcff1a110</wsa:To>
    <wsa:ReplyTo>
      <wsa:Address>http://schemas.xmlsoap.org/ws/2004/08/addressing/role/anonymous</wsa:Address>
    </wsa:ReplyTo>
  </s12:Header>
  <s12:Body />
</s12:Envelope>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T02:43:59+00:00Added an answer on May 14, 2026 at 2:43 am
    1. First, you can create a String template for this SOAP request and substitute user-supplied values at runtime in this template to create a valid request.
    2. Wrap this string in a StringEntity and set its content type as text/xml
    3. Set this entity in the SOAP request.

    Something like:

    HttpPost httppost = new HttpPost(SERVICE_EPR);          
    StringEntity se = new StringEntity(SOAPRequestXML,HTTP.UTF_8);
    
    se.setContentType("text/xml");  
    httppost.setHeader("Content-Type","application/soap+xml;charset=UTF-8");
    httppost.setEntity(se);  
    
    HttpClient httpclient = new DefaultHttpClient();
    BasicHttpResponse httpResponse = 
        (BasicHttpResponse) httpclient.execute(httppost);
    
    response.put("HTTPStatus",httpResponse.getStatusLine().toString());
    
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