android soapfault 服务器无法处理请求,使用kso​​ap2的WCF Web服务的Android – 错误SoapFault – faultcode：’a：InternalServic...

我做了一个简单的项目,使用kso​​ap2调用wcf Web服务.但是当它调用envelope.getResponse();它给出了错误的说法

错误：

SoapFault – faultcode：’a：InternalServiceFault’faultstring：’由于内部错误,服务器无法处理请求.有关错误的更多信息,请在服务器上启用IncludeExceptionDetailInFaults(来自ServiceBehaviorAttribute或配置行为)以将异常信息发送回客户端,或者根据Microsoft .NET Framework 3.0 SDK文档启用跟踪并检查服务器跟踪日志. faultactor：’null’detail：null

package testing.wcf;

import java.io.IOException;

import org.ksoap2.SoapEnvelope;

import org.ksoap2.serialization.SoapObject;

import org.ksoap2.serialization.SoapPrimitive;

import org.ksoap2.serialization.SoapSerializationEnvelope;

import org.ksoap2.transport.HttpTransportSE;

import org.xmlpull.v1.XmlPullParserException;

import org.xmlpull.v1.XmlSerializer;

import android.os.Bundle;

import android.os.StrictMode;

import android.widget.TextView;

import android.annotation.SuppressLint;

import android.app.Activity;

public class MainActivity extends Activity

{

private static final String strNAMESPACE = "http://www.testing.co.in/TestingService/";

private static final String strURL = "http://www.testing.co.in/TestingService/UserDetails.svc";

private static final String strSOAP_ACTION = "http://testing.co.in/TestingService/UserDetails/LoginDetails";

private static final String strMETHOD_NAME = "LoginDetails";

TextView tv;

StringBuilder sb;

String strInputXML;

@Override

public void onCreate(Bundle savedInstanceState)

{

super.onCreate(savedInstanceState);

setContentView(R.layout.activity_main);

tv = (TextView)findViewById(R.id.testing);

sb = new StringBuilder();

Call();

tv.setText(sb.toString());

//setContentView(tv);

}

public void Call()

{

try

{

SoapObject request = new SoapObject(strNAMESPACE, strMETHOD_NAME);

String inputxml = "<?xml version="+"""+"1.0"+"""+" encoding="+"""+"utf-8"+"""+" ?>" +" n 0000 n TEST n 112233 n LoginVal n ";

request.addAttribute("strInputXML", inputxml);

request.addAttribute("strOutputXML","");

SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);

envelope.dotNet = true;

envelope.setOutputSoapObject(request);

HttpTransportSE androidHttpTransport = new HttpTransportSE(strURL);

androidHttpTransport.call(strSOAP_ACTION, envelope);

SoapPrimitive result = (SoapPrimitive)envelope.getResponse();

String resultData = result.toString();

sb.append(resultData + "n");

}

catch(Exception e)

{

sb.append("Error:n" + e.getMessage() + "n");

}

}

}

在这里,我想发送这样的请求

0000

HELLO

1234

0

我的回应应该是XML

1234