Android SOAP Web Service - 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。肥皂对象响应时出错。

我尝试使用肥皂原语，但不起作用。请帮助我问题所在

给定的 Web 服务返回 xml。

Answer 1

cityInfo.setValue(city);

您在代码中的何处获取此"城市"值？我认为这就是问题所在。为了进行测试，您可以直接在此处将其设置为字符串，例如"纽约"。