android soap Web服務-不檢索輸出

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();
                      }

顯示空異常。 soapObject響應發生錯誤。

我嘗試使用原始肥皂，但無法正常工作。 請幫我解決問題所在

給定的Web服務返回xml。

Answer 1

cityInfo.setValue(city);

您在哪里從代碼中獲取此“城市”值？ 我認為那是問題。 為了進行測試，您可以在此處直接將其設置為字符串，例如“ NewYork”。