在Android中使用kso​​ap2调用.NET Web服务，没有返回结果

Question

我一直试图通过android使用kso​​ap2调用一个简单的.NET webservice（HelloWorld）（我已经尝试并成功使用了不同的webservice）。但是这个需要身份验证，所以我搜索了如何为身份验证添加标头，但是，我的textview中没有返回任何结果。

SoapObject Request = new SoapObject(NAMESPACE,METHOD_NAME);  
List<HeaderProperty> headers = new ArrayList<HeaderProperty>();
headers.add(new HeaderProperty("Authorization", "Basic"+Base64.encode("Username:Password".getBytes())));


        SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
        envelope.dotNet=true;
        envelope.setOutputSoapObject(Request);

        HttpTransportSE aht = new HttpTransportSE(URL);


        try{
            aht.call(SOAP_ACTION,envelope,headers);
            SoapPrimitive resultString = (SoapPrimitive)envelope.getResponse();

            tv.setText("yo :" + resultString);

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

我的日志中也有错误：

  

错误：线程附加失败，但我认为它不是

的来源

这是HelloWorld方法的wsdl：

  

>     <wsdl:types>
>     <s:schema elementFormDefault="qualified" targetNamespace="http://tempuri.org/">
>     <s:element name="HelloWorld">
>     <s:complexType/>
>     </s:element>
>     <s:element name="HelloWorldResponse">
>     <s:complexType>
>     <s:sequence>
>     <s:element minOccurs="0" maxOccurs="1" name="HelloWorldResult" type="s:string"/>
>     </s:sequence>
>     </s:complexType>
>     </s:element>

  

我的问题。   有人遇到过这种问题，或者我的代码在某些时候出错了吗？

Answer 1

试试这个：

 SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(
                        SoapEnvelope.VER11);    
request.addProperty("requst_name",request_value);//if you have any request add here..                   
 envelope.setOutputSoapObject(request);
 envelope.implicitTypes = true;
 envelope1.dotNet = true;
 int Timeout = 60000;
 HttpTransportSE androidHttpTransport = new HttpTransportSE(URL, Timeout);
 androidHttpTransport.debug = true;
                try {

                    androidHttpTransport.call(SOAP_ACTION, envelope);
                    Object response = envelope.getResponse();
                    String res=response.toString();
                    }

                    }catch (XmlPullParserException e) {

                        e.printStackTrace();
                    } catch (SocketTimeoutException e) {
                        // TODO Auto-generated catch block
                        e.printStackTrace();
                    } catch (IOException e) {
                        // TODO Auto-generated catch block
                        e.printStackTrace();
                    }

Answer 2

。这就是我访问.net网络服务的方式。在我的情况下，我发送2个参数用户名和密码，服务将根据数据库返回一个字符串。如果你没有传递任何参数请删除我提到的参数，只添加这个

PropertyInfo pi = new PropertyInfo();
    pi.setType(String.class);
    request.addProperty(pi);

---

否则

 public String Call(String username, String password) {
    SoapObject request = new SoapObject(WSDL_TARGET_NAMESPACE,
            OPERATION_NAME);

    PropertyInfo pi = new PropertyInfo();
    pi.setName("username");
    pi.setValue(username);
    pi.setType(String.class);
    request.addProperty(pi);

    pi = new PropertyInfo();
    pi.setName("password");
    pi.setValue(password);
    pi.setType(String.class);
    request.addProperty(pi);

    SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(
            SoapEnvelope.VER11);

    envelope.dotNet = true;

    envelope.setOutputSoapObject(request);

    HttpTransportSE httpTransport = new HttpTransportSE(SOAP_ADDRESS);
    Object response = null;

    try {

        httpTransport.call(SOAP_ACTION, envelope);
        response = envelope.getResponse();
    }

    catch (Exception exception) {
        response = exception.toString();
    }

    return response.toString();
}