I'm trying to access a Java Web Service from Android client, but it is showing me an error:
"java.lang.classcastexception org.ksoap2.soapfault cannot be cast to org.ksoap2.serialization.soapobject"
Can you help me?
Here is my client web service code:
import java.lang.reflect.Method;
import android.app.Activity;
import android.os.Bundle;
import android.content.Context;
import android.content.Intent;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.view.Window;
import android.widget.EditText;
import android.widget.TextView;
import org.ksoap2.SoapEnvelope;
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.AndroidHttpTransport;
import org.ksoap2.transport.HttpTransportSE;
public class Loginuser extends Activity{
public static final int MENU1 = Menu.FIRST;
public static final int MENU2 = Menu.FIRST + 1;
public static final int MENU3 = Menu.FIRST + 2;
public static Context group;
private static final String SOAP_ACTION = "";
private static final String METHOD_NAME = "logar";
private static final String NAMESPACE = "http://wsproj.mycompany.com/";
private static final String URL = "http://localhost:8084/wsproj/HelloWorld";
EditText ura,pw;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
requestWindowFeature(Window.FEATURE_NO_TITLE);
setContentView(R.layout.loginuser);
}
public void logar(View X) {
CarregaTelaBolarq();
}
public void CarregaTelaBolarq(){
ura=(EditText)findViewById(R.id.editText2);
String raforn = ura.getText().toString();
SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
request.addProperty("raforn",ura.getText().toString());
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.setOutputSoapObject(request);
try{
HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
androidHttpTransport.call(SOAP_ACTION, envelope);
SoapObject sp = (SoapObject)envelope.bodyIn;
String result=sp.toString();
if(result.equals("1"))
{
TextView tv;
tv=(TextView) findViewById(R.id.editText1);
tv.setText("foi: ");
}
else
{
TextView tv;
tv=(TextView) findViewById(R.id.editText1);
tv.setText("Msg from service: ");
}
}
catch(Exception e)
{
TextView tv=(TextView) findViewById(R.id.editText1);
tv.setText("ERROR: " + e.toString());
}
}
public boolean onCreateOptionsMenu(Menu options) {
options.add(0, MENU1, 0, "Página Principal");
options.add(0, MENU2, 0, "Manual");
options.add(0, MENU3, 0, "Sobre");
return super.onCreateOptionsMenu(options); }
public boolean onOptionsItemSelected(MenuItem item) {
switch (item.getItemId()) {
case MENU1:
Intent mudarHome= new Intent(this, MainActivity.class);
startActivity(mudarHome);
return true;
case MENU2:
Intent mudarManual = new Intent(this, Manual.class);
startActivity(mudarManual);
return true;
case MENU3:
Intent mudarSobre = new Intent(this, Sobre.class);
startActivity(mudarSobre);
return true;
}
return false;
}
}
That's means there is no service found by those parameters try this code to find the error message :
SoapFault error = (SoapFault)envelope.bodyIn;
System.out.println("Error message : "+error.toString());
in my opinion you must fill the SOAP_ACTION parameter by the class that include the service with the package name :
private static final String SOAP_ACTION = "http://com.mycompany.wsproj/HelloWorld";
and end the URL of the web service by .wsdl or ?wsdl ( try them both xD )
private static final String URL = "http://localhost:8084/wsproj/HelloWorld?wsdl";
one last important thing is (when you are using android API ) change the localhost by the IP :
private static final String URL = "http://10.0.2.2:8084/wsproj/HelloWorld?wsdl";
Hope that helps you !! ... good luck !