2009-06-26 133 views
26

我需要调用一个网址为:"http://192.168.1.19/TestWeb/WebService.asmx"如何从android调用.NET Web服务?

任何人都可以帮我一个完整的例子吗?

+0

我不明白。您的网络服务使用.NET。 Android进入它的哪里? HelloWorld方法在哪里? – 2009-06-26 10:08:39

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感谢您的编辑。你试过什么了?你有什么问题? – 2009-06-26 12:15:17

回答

31

最后,我得到了我自己的问题的解决方案。

下面是代码:

package projects.ksoap2sample; 



import org.ksoap2.SoapEnvelope; 
import org.ksoap2.serialization.SoapObject; 
import org.ksoap2.serialization.SoapSerializationEnvelope; 
import org.ksoap2.transport.HttpTransportSE; 
import android.app.*; 
import android.os.*; 
import android.widget.TextView; 

public class ksoap2sample extends Activity { 
    /** Called when the activity is first created. */ 
    private static final String SOAP_ACTION = "http://tempuri.org/HelloWorld"; 

    private static final String METHOD_NAME = "HelloWorld"; 

    private static final String NAMESPACE = "http://tempuri.org/"; 
    private static final String URL = "http://192.168.1.19/TestWeb/WebService.asmx"; 
    TextView tv; 

    @Override 
    public void onCreate(Bundle savedInstanceState) { 
     super.onCreate(savedInstanceState); 
     setContentView(R.layout.main); 
     tv=(TextView)findViewById(R.id.text1); 
     call(); 

    } 

    public void call() 
    { 
     try { 

      SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME); 

      request.addProperty("passonString", "Rajapandian"); 

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

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

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

      tv.setText(result.toString()); 
     } catch (Exception e) { 
      tv.setText(e.getMessage()); 
      } 
    } 
} 

问候 Rajapandian