2011-10-11 88 views
0

我需要幫助。我試圖將參數傳遞給webservice,它顯示我有錯誤: CAN NOT SERIALIZE content_of_userText。 我的代碼是:無法在Android上序列化錯誤

if (button.getId() == R.id.loginButton) { 

      //Modelo el request 
      SoapObject request = new SoapObject(namespace, Metodo); 

      //Paràmetres 
      request.addProperty("correu", userText.getText()); 
      request.addProperty("pass", passText.getText()); 

      //Modelo el Sobre 
      SoapSerializationEnvelope sobre = new SoapSerializationEnvelope(SoapEnvelope.VER11); 
      sobre.dotNet = true; 
      sobre.setOutputSoapObject(request); 

      //Modelo el transporte 
      HttpTransportSE transporte = new HttpTransportSE(url); 

      try {  
       //Llamada 
       transporte.call(accionSoap, sobre); 

       //Resultado 
       SoapObject resultado = (SoapObject) sobre.getResponse(); 

       Toast.makeText(this, resultado.toString(), Toast.LENGTH_SHORT).show(); 

      } 
      catch (Exception e) 
      { 
       Toast.makeText(this, e.getMessage(), Toast.LENGTH_SHORT).show(); 
      } 

謝謝!!

編輯:我試試這個:

request.addProperty( 「correu」,userText.getText()的toString());
request.addProperty(「pass」,passText.getText()。toString());

和例外現在是:

org.ksoap2.serialization.SoapPrimitive

EDIT2:它的工作原理!我改變了:

SoapObject resultado = (SoapObject) sobre.getResponse(); 

此:

String resultado = sobre.getResponse().toString(); 

回答

0

試試這個

request.addProperty("correu", userText.getText().toString()); 
request.addProperty("pass", passText.getText().toString()); 

而不是

request.addProperty("correu", userText.getText()); 
    request.addProperty("pass", passText.getText());