Я использую в своем коде веб-сервис asmx, который не возвращает правильный XML-файл.Вот мой код
private static final String SOAP_ACTION = "http://tempuri.org/ValidateLogin";
private static final String METHOD_NAME = "ValidateLogin" ;
private static final String NAMESPACE = "http://tempuri.org/";
private static final String URL = "http://ufindfish.b4live.com/FindFish.asmx";
request= new SoapObject(NAMESPACE,METHOD_NAME);
request.addProperty("sUserName",name);
request.addProperty("sPwd", password);
//envelope.bodyOut=request;
envelope=new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet=true;
envelope.setOutputSoapObject(request);
envelope.encodingStyle=SoapSerializationEnvelope.XSD;
//generate httpresponce
httpTransportSE=new HttpTransportSE(URL);
try {
httpTransportSE.call(SOAP_ACTION,envelope);
} catch (IOException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
} catch (XmlPullParserException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
SoapObject result = null;
try {
result=(SoapObject)envelope.getResponse();
//Log.i("RESPONCE",""+result.toString());
} catch (SoapFault e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
ответ ниже ... что мне делать?
anyType{NearLake=anyType{tblNearLake=anyType{LAKE=2022C01 BENSFORT BRIDGE to PETERBOROUGH; LAT=11.622; LON=20.616; }; tblNearLake=anyType{LAKE=2022C01 BENSFORT BRIDGE to PETERBOROUGH; LAT=11.186; LON=19.443; }; }; }
помогите пожалуйста ...
Заранее спасибо