Android：肥皂原始错误

我在访问Web服务时遇到了一些问题。获取ClassCastException错误。考虑一个场景，我试图访问一个Web服务的方法，Web服务应该返回两个字符串（比如说String1和String2）。而且，我必须提供或传递两个参数（可以说参数1和参数2，其中参数1应该是整数和参数2应该是字符串）这里是我的代码Android：肥皂原始错误

public class MyWebService extends Activity { 

private static final String SOAP_ACTION ="http://www.mywebsite.com/myMethod"; 
private static final String METHOD_NAME = "MyMethod"; 
private static final String NAMESPACE = "http://www.myNamespace/"; 
    private static final String URL = "http://mysession.com/myservice.asmx?WSDL"; 


@Override 
public void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.main); 

    try { 
     SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME); 

     PropertyInfo pi = new PropertyInfo(); 
     pi.setName("Parameter 1"); 
     pi.setValue(1); 
     pi.setType(pi.INTEGER_CLASS); 
     request.addProperty(pi); 

     PropertyInfo pi2 = new PropertyInfo(); 
     pi2.setName("Parameter 2"); 
     pi2.setValue("Any string"); 
     pi2.setType(pi2.STRING_CLASS); 
     request.addProperty(pi2); 

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

     HttpTransportSE androidHttpTransport = new HttpTransportSE(URL); 
     androidHttpTransport.call(SOAP_ACTION, envelope);     
     SoapObject result=(SoapObject)envelope.getResponse();    


     String string1=result.getProperty(0).toString(); 
     String string2=result.getProperty(1).toString(); 


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

    }   
}

}

这是除了我得到 java.lang.ClassCastException：org.ksoap2.serialization.SoapPrimitive

谁能告诉我，如果我错了，在这里做什么.. 感谢

来源

2011-09-23 Farhan

试试这个，

SoapPrimitive result= (SoapPrimitive)envelope.getResponse();

Object result= (Object)envelope.getResponse();

，而不是

SoapObject result=(SoapObject)envelope.getResponse();

来源

2011-09-24 03:00:26

-1

试试这个：

SoapObject request = new SoapObject(SOAP_NAMESPACE,SOAP_METHOD); 
request.addProperty("name", abcd); 
request.addProperty("age", 30);

来源

2016-05-12 08:28:41

作为一个大师，你能解释多一点有什么不好，为什么你认为他应该尝试这个？ –

Android：肥皂原始错误

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