How to access Webservice in android?

Question

I am new in android development, i need to get the XML file from the Webservice by using SOAP message. I have tried my level best to find out SOAP message parsing in Android but, i cannot find out exact solution for the SOAP message parsing. Here i have attached my sample code to parse the SOAP message. Can you please help me to parse the SOAP message in Android? (Response comes in permission denied).

I tried source code given below:

SoapObject request = new SoapObject(NAMESPACE ,METHOD_NAME);
request.addProperty("username","d");
request.addProperty("password","d123");
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet=true;
envelope.setOutputSoapObject(request);envelope.setAddAdornments(true);
HttpTransportSE httpTransport = new HttpTransportSE(URL);  
try 
  {
      httpTransport.call(SOAP_ACTION, envelope); //send request
      SoapObject result=(SoapObject)envelope.bodyIn;
      String results = result.toString();
      tv.setText( ""+results);
  }
  catch (Exception e)
  {
       tv.setText(e.getMessage());
   }

AndroidManifest.Xml file:

<uses-permission android:name="android.permission.INTERNET">
</uses-permission>

Is REST not an option, also make sure you do this in a background thread separate to the main UI thread like I explained in another question I answered yesterday here stackoverflow.com/questions/8812281/… — Graham Smith
– Graham Smith, Commented Jan 11, 2012 at 10:23

Lalit Poptani · Accepted Answer · 2012-01-11 11:24:54Z

1

You are missing envelope.getResponse(); in your code, prabably that can be the reason that you are not getting the Response.

You can try,

SoapObject result=(SoapObject)envelope.getResponse();
String results = result.toString();

edited Jan 11, 2012 at 11:24

answered Jan 11, 2012 at 11:17

Lalit Poptani

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2 Comments

Sampath Kumar Over a year ago

Thanks you, but response is same permission denied

Lalit Poptani Over a year ago

Post the exception in the question.

sudeepjha · Accepted Answer · 2012-01-12 10:20:02Z

0

You can do it like this

        HttpPost httppost = new HttpPost(webServicePath);
    httppost.setHeader("Content-Type", "text/xml;charset=UTF-8");

    HttpParams params = new BasicHttpParams();
    params.setParameter(CoreProtocolPNames.PROTOCOL_VERSION,
            HttpVersion.HTTP_1_1);

    String soapRequestXML = getXMLAsString();
    soapRequestXML = prepareInputParam(soapRequestXML);

    StringEntity se;
    se = new StringEntity(soapRequestXML, HTTP.UTF_8);
    se.setContentType("text/xml");
    httppost.setEntity(se);

    response = httpClient.execute(httppost);

answered Jan 12, 2012 at 10:20

sudeepjha

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