For .net services, soap action should be SOAPAction: "http://tempuri.org/CelsiusToFahrenheit";
check it out from http://www.w3schools.com/webservices/tempconvert.asmx?op=CelsiusToFahrenheit through your web browser Alan On Thu, 17 Dec 2009 06:24:19 -0800 (PST), android09 <[email protected]> wrote: > Hello Everyone, > > I am trying to develop an application which consume the asp.net web > service from android. I tried a lot but every time i got an exception > while calling the SOAP_ACTION. I have tried with the localhost, local > IP Address and also i have taken URL from the internet but it throws > an exception. Here it is: > > Exception: > ------------------ > SoapFault - faultcode: 'soap:Client' faultstring: 'Server did not > recognize the value of HTTP HeaderSOAPAction : > http://www.w3schools.com/webservices/tempconvert.asmx?op=CelsiusToFahrenheit.' > faultactor: 'null' detail : org.kxml2.kdom.n...@43c582e8 > > I also got an exception like this:- "org.ksoap2.SoapFault : null" > > Here is my source code: > ------------------------------------ > package com.android.webdemo3; > > import org.ksoap2.SoapEnvelope; > import org.ksoap2.serialization.SoapObject; > import org.ksoap2.serialization.SoapSerializationEnvelope; > import org.ksoap2.transport.HttpTransportSE; > > import android.app.*; > import android.os.*; > import android.widget.TextView; > > public class WebDemo3 extends Activity { > /** Called when the activity is first created. */ > > private static final String SOAP_ACTION = "http://www.w3schools.com/ > webservices/tempconvert.asmx?op=CelsiusToFahrenheit"; > private static final String METHOD_NAME = "CelsiusToFahrenheit"; > private static final String NAMESPACE = "http://tempuri.org/ > CelsiusToFahrenheit"; > private static final String URL = "http://www.w3schools.com/ > webservices/tempconvert.asmx"; > > TextView tv; > > @Override > public void onCreate(Bundle savedInstanceState) { > super.onCreate(savedInstanceState); > setContentView(R.layout.main); > tv=(TextView)findViewById(R.id.textView); > > try > { > SoapObject request = new SoapObject(NAMESPACE, > METHOD_NAME); > > // request.addProperty("Celsius", 32); > > SoapSerializationEnvelope envelope = new > SoapSerializationEnvelope(SoapEnvelope.VER11); > envelope.dotNet=true; > envelope.setOutputSoapObject(request); > > HttpTransportSE androidHttpTransport = new HttpTransportSE > (URL); > > androidHttpTransport.call(SOAP_ACTION, envelope);//Here an > exception is throw. > > // Object result = (Object)envelope.getResponse(); > // String[] results = (String[]) result; > // System.out.println("Result is: "); > > java.lang.String receivedString = (String) > envelope.getResponse(); > tv.setText("Result is: "+receivedString); > } > catch (Exception e) > { > e.printStackTrace(); > tv.setText(e.getMessage()); > } > } > } > > You can access this URL, because i got from the internet(web). I can't > understand, why this exception is throw. Please anybody give me > solution of my problem and i would really appreciate your help. Please > reply me as soon as possible earlier. > > Thanks. > Pranav -- You received this message because you are subscribed to the Google Groups "Android Developers" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/android-developers?hl=en
