Question

我需要使用以下URL来调用Web服务："http://192.168.1.19/TestWeb/WebService.asmx"来自android。

请有人帮我一个完整的例子吗？

Answer 1

最后，我得到了自己问题的解决方案。

以下是代码：

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());
            }
    }
}

此致 Rajapandian

如何从android调用.NET Web服务？

1 个答案: