獲取任何類型{} fron soapobject

我從soapobject獲取任何類型的響應。{}我正在嘗試使用Web服務傳遞用戶名和密碼，但無法獲得相同的用戶名和密碼。請任何身體幫助。獲取任何類型{} fron soapobject

private static final String NAMESPACE = "http://tempuri.org/"; 
private static final String URL = "http://***.***.*.*/MObile/Logics.asmx"; 
private static final String LOGIN_METHOD = "CheckLogin"; 
private static final String SOAP_ACTION_LOGIN = "http://tempuri.org/CheckLogin"; 


@Override 
public void onCreate(Bundle savedInstanceState) 
{ 
    super.onCreate(savedInstanceState); 
    StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build(); 
    StrictMode.setThreadPolicy(policy); 
    setContentView(R.layout.login_layout); 

    Button signin = (Button) findViewById(R.id.btn_login); 
    signin.setOnClickListener(this); 

} 
     public void onClick(View v) { 

      try { 

      etxt_user = (EditText) findViewById(R.id.txt_username); 
      etxt_password = (EditText) findViewById(R.id.txt_password); 

      String username = etxt_user.getText().toString(); 
      String password = etxt_password.getText().toString(); 

      result = (TextView)findViewById(R.id.tv); 

      SoapObject request = new SoapObject(NAMESPACE, LOGIN_METHOD); 

      request.addProperty("username",username); 
      request.addProperty("password",password); 

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

      Log.i("LoginDetail", "Username " + username + "Password " + password); 

      HttpTransportSE androidHttpTransport = new HttpTransportSE(URL); 
      androidHttpTransport.debug = true; 

       androidHttpTransport.call(SOAP_ACTION_LOGIN, envelope); 

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

       Log.i("OUTPUT", resultString.toString()); 
       // getting output anytype{} 

       if(){ 

        } 

        else{ 

         Toast.makeText(getApplicationContext(), "Wrong U & P", Toast.LENGTH_SHORT).show(); 

        }*/ 

      } 
      catch (Exception e) { 

       e.printStackTrace(); 
       Toast.makeText(getApplicationContext(), " Network Exception : " + e 
           + "Please check network connectivity.", Toast.LENGTH_LONG).show(); 

      } 

     }

}

我堅持這個問題。我想要如果用戶名密碼正確移動到下一個活動，如果錯誤，然後toast.show（）。

如何驗證用戶名爲&的密碼是否正確？

來源

2013-02-06 Shweta

anytype類型{}是webservice.So空值迴歸ü可以檢查您的WS？ –

是的，謝謝。我得到了答案。 – Shweta

試着在後面加上下面一行：

envelope.implicitTypes= false;

後

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

來源

2013-02-06 11:07:36 TheWhiteRabbit

獲取任何類型{} fron soapobject

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