XmlPullParser 异常

Question

我无法解决此错误：

org.xmlpull.v1.XmlPullParserException：预期：START_TAG {http://www. w3.org/2001/12/soap-envelope}信封（位置：START_TAG @1:6 in java.io.InputStreamReader@43e524c8）

我有一个访问 .NET Web 服务的 Android 应用程序。它传递一个 int 并接收一个 int 作为响应。我正在使用 ksoap2。 我还向清单文件添加了权限，而且还包含了所有 jar 文件。但是，我陷入困境，无法查看我正在发送的 SOAP 请求。

我的代码是：

public class FirstAppUI extends Activity{

    private static final String NAMESPACE = "http://tempuri.org/";

    private static final String URL = 
        "http://nautilussoft.biz.whbus12.onlyfordemo.com/staging/litigation/demowebservice.asmx";

private static final String SOAP_ACTION ="http://tempuri.org/GetNumber";

private static final String METHOD_NAME = "GetNumber";

    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        TextView tv= (TextView)findViewById(R.id.TextView01);
        SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
        request.addProperty("maxbelow", "123");
        SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11); 

        envelope.dotNet= true;
        envelope.setOutputSoapObject(request);
        HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
        try {
            //Toast.makeText(getBaseContext(), "received object", Toast.LENGTH_SHORT).show();
            androidHttpTransport.call(SOAP_ACTION, envelope);

             Object resultsRequestSOAP = (SoapObject)envelope.getResponse();
            //SoapObject resultsRequestSOAP = (SoapObject) envelope.bodyIn;
            //Toast.makeText(getBaseContext(), "received object", Toast.LENGTH_SHORT).show();
            tv.setText("Received :" + resultsRequestSOAP.toString());
        } catch (MalformedURLException e) {
            e.printStackTrace();
            Log.e("APP", "MalformedURLException while sending\n" + e.getMessage());
            tv.setText("Malformedexception"+e);
        }
        catch(Exception e1)
        {
            e1.printStackTrace();
            tv.setText("exception"+e1);

        }

        }

}

提前致谢。

Answer 1

在 IIS 中或通过 IIS 管理器授予对 webserivce 文件夹的权限。

Answer 2

请检查您正在使用的 weburl...它区分大小写...即使它在网络浏览器中工作正常，您也应该在代码中使用确切的名称...