我无法解决此错误：
org.xmlpull.v1.XmlPullParserException：预期：START_TAG { http://www.w3.org/2001/12/soap-envelope }信封（位置：START_TAG @1:6 in java.io.InputStreamReader@43e524c8）
我有一个访问 .NET Web 服务的 Android 应用程序。它传递一个 int 并接收一个 int 作为响应。我正在使用 ksoap2。我还向清单文件添加了权限，而且我包含了所有 jar 文件。但是，我被卡住了，无法查看我正在发送的 SOAP 请求。
我的代码是：
public class FirstAppUI extends Activity{
private static final String NAMESPACE = "http://tempuri.org/";
private static final String URL =
"http://nautilussoft.biz.whbus12.onlyfordemo.com/staging/litigation/demowebservice.asmx";
private static final String SOAP_ACTION ="http://tempuri.org/GetNumber";
private static final String METHOD_NAME = "GetNumber";
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
TextView tv= (TextView)findViewById(R.id.TextView01);
SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
request.addProperty("maxbelow", "123");
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet= true;
envelope.setOutputSoapObject(request);
HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
try {
//Toast.makeText(getBaseContext(), "received object", Toast.LENGTH_SHORT).show();
androidHttpTransport.call(SOAP_ACTION, envelope);
Object resultsRequestSOAP = (SoapObject)envelope.getResponse();
//SoapObject resultsRequestSOAP = (SoapObject) envelope.bodyIn;
//Toast.makeText(getBaseContext(), "received object", Toast.LENGTH_SHORT).show();
tv.setText("Received :" + resultsRequestSOAP.toString());
} catch (MalformedURLException e) {
e.printStackTrace();
Log.e("APP", "MalformedURLException while sending\n" + e.getMessage());
tv.setText("Malformedexception"+e);
}
catch(Exception e1)
{
e1.printStackTrace();
tv.setText("exception"+e1);
}
}
}
提前致谢。