android soap Web服务-不检索输出

Question

private final String zipCodenameSpace ="http://www.webserviceX.NET/";
private final String zipURL="http://www.webserviceX.net/uszip.asmx";
private final String zipSoapAction ="http://www.webserviceX.NET/GetInfoByCity";
private final String zipMethodName="GetInfoByCity";



SoapObject request = new SoapObject(zipCodenameSpace, zipMethodName);
         PropertyInfo cityInfo = new PropertyInfo();
         cityInfo.setName("USCity");
         cityInfo.setValue(city);
        // cityInfo.setType(String.class);
         SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
         request.addProperty(cityInfo);
         TextView tv = (TextView) findViewById(R.id.tv1);

         envelope.dotNet = true;

         envelope.setOutputSoapObject(request);
         HttpTransportSE androidHttpTransport = new HttpTransportSE(zipURL);

         try {
             androidHttpTransport.call(zipSoapAction, envelope);

         //    Object response =(SoapObject) envelope.getResponse();
            SoapObject response = (SoapObject) envelope.getResponse();

            Log.d("This is an element", response.toString());
          //   tv.setText(response.toString());

           //  return response.toString();

          } catch (Exception e) {
             e.printStackTrace();
                      }

显示空异常。 soapObject响应发生错误。

我尝试使用原始肥皂，但无法正常工作。 请帮我解决问题所在

给定的Web服务返回xml。

Answer 1

cityInfo.setValue(city);

您在哪里从代码中获取此“城市”值？ 我认为那是问题。 为了进行测试，您可以在此处直接将其设置为字符串，例如“ NewYork”。