Android KSoap2没有返回有效的xml?

Android KSoap2没有返回有效的xml?,android,xml,soap,android-ksoap2,Android,Xml,Soap,Android Ksoap2,我正在尝试从web服务获取xml数据。xml数据如下所示 <Result> <ErrorCode>0</ErrorCode> <ErrorMessage>Login was succesful.</ErrorMessage> <AuthCode>maneen90234</AuthCode> </Result> 这就是我调用web服务的方式 private String callAPI(String

我正在尝试从web服务获取xml数据。xml数据如下所示

<Result>
<ErrorCode>0</ErrorCode>
<ErrorMessage>Login was succesful.</ErrorMessage>
<AuthCode>maneen90234</AuthCode>
</Result>
这就是我调用web服务的方式

private String callAPI(String user, String password) {
        request = new SoapObject(NAME_SPACE, "UserLogin");
        PropertyInfo pi = new PropertyInfo();
        addProperty(pi, request, "UserId", user, String.class);
        addProperty(pi, request, "Password", password, String.class);
        envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
        envelope.dotNet = true;

        envelope.setOutputSoapObject(request);

        androidHttpTransport = new HttpTransportSE(url);

        try {
            androidHttpTransport.call(SOAP_ACTION, envelope);
            response = envelope.getResponse().toString();
        } catch (Exception exception) {
            response = exception.toString();

        }
        return response;
    }

    private void addProperty(PropertyInfo pi, SoapObject request, String UserId, String value, Object type) {
        pi = new PropertyInfo();
        pi.setName(UserId);
        pi.setValue(value);
        pi.setType(type);
        request.addProperty(pi);
    }

我试过各种各样的答案,但都没有用。知道我做错了什么吗?

看起来ksoap2正在解析数据本身并返回数据的属性。我提取的数据如下所示-

SoapObject response = (SoapObject) ((SoapObject)envelope.getResponse()).getProperty(0);
 if(response.getProperty("ErrorCode").equals(1))
        {
            Toast.makeText(context,"Login sucessful",Toast.LENGTH_LONG).show();
            Log.d(TAG, String.valueOf(response));
        }
        else {
            Toast.makeText(context,"ErrorCode",Toast.LENGTH_LONG).show();
            Log.d(TAG, String.valueOf(response));
        }

虽然我仍然在传递参数方面遇到问题,但我将针对相同的问题发布单独的问题。

当你说“ErrorMessage=不正确的用户名或密码;”@sakir是的,我意识到我得到的是已解析的数据,不需要再次解析。我希望有一个有效的XML作为响应。请检查我的答案以了解解释。
SoapObject response = (SoapObject) ((SoapObject)envelope.getResponse()).getProperty(0);
 if(response.getProperty("ErrorCode").equals(1))
        {
            Toast.makeText(context,"Login sucessful",Toast.LENGTH_LONG).show();
            Log.d(TAG, String.valueOf(response));
        }
        else {
            Toast.makeText(context,"ErrorCode",Toast.LENGTH_LONG).show();
            Log.d(TAG, String.valueOf(response));
        }