Question

我创建了一个示例Android应用程序，即调用简单的.net webservice并在TextView中显示，下面是代码前景：

网络服务代码：

[WebService(Namespace = "http://tempuri.org/")]
[WebServiceBinding(ConformsTo = WsiProfiles.BasicProfile1_1)]
// To allow this Web Service to be called from script, using ASP.NET AJAX, uncomment the following line. 
 [System.Web.Script.Services.ScriptService]
public class Service : System.Web.Services.WebService
{
    public Service () {

        //Uncomment the following line if using designed components 
        //InitializeComponent(); 
    }

    [WebMethod]
    public string MobileApp(string acNum) 
    {
        return "Account number is"+acNum;
    }

}

android app code：

import java.io.IOException;

import android.app.TabActivity;
import android.content.Intent;
import android.content.res.Resources;
import android.os.Bundle;
import android.view.View;
import android.widget.TabHost;
import android.widget.TextView;

import org.ksoap2.SoapEnvelope;
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.HttpTransportSE;
import org.xmlpull.v1.XmlPullParserException;


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

    private static final String METHOD_NAME = "MobileApp";

    private static final String NAMESPACE = "http://tempuri.org/";
    private static final String URL = "http://localhost:58817/MobService/Service.asmx";
    private TextView tv;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.accountdet);
        tv=(TextView)findViewById(R.id.tView);
ServiceCall();
    }
    public void ServiceCall()
    {
        try{
        SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
        HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
        request.addProperty("acNum", "123456");

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


        androidHttpTransport.call(SOAP_ACTION, envelope);

        SoapObject result = (SoapObject)envelope.getResponse();
        tv.setText(result.toString());

        }       
        catch (final IOException e)

        {

            // TODO Auto-generated catch block

            e.printStackTrace();

        } catch (final XmlPullParserException e)

        {

            // TODO Auto-generated catch block

            e.printStackTrace();

        } catch (final Exception e)

        {

            // TODO Auto-generated catch block

            e.printStackTrace();

        }


    }

此外，我在清单文件中添加了以下行：

<uses-permission android:name="android.permission.INTERNET" ></uses-permission>
<uses-sdk android:minSdkVersion="7" />

但是我没有在TextView中得到结果，所以任何人都可以帮助我。

@nagaraju。

Answer 1

首先检查您的服务是否正在运行 - 为此尝试在浏览器中显示您的链接link，然后创建虚拟网站，在那里调用您的方法MobileApp，如果它返回您的服务工作正常。

  SoapSerializationEnvelope envelopes = new SoapSerializationEnvelope(SoapEnvelope.VER11); // put all required data into a soap//// envelope
    envelopes.dotNet = true;
    envelopes.setOutputSoapObject(requestData);
    AndroidHttpTransport httpTransport = new AndroidHttpTransport(APPURL);
    httpTransport.debug = true;

    try {
        httpTransport.call(SOAP_ACTION1, envelopes);
        responsesData = (SoapPrimitive) envelopes.getResponse();

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

试试这个..

Answer 2

我发现了我的错误，很简单，因为我在应用程序标记的清单文件中放了用户权限行，但实际上，我要放在应用程序标记之外，这就是为什么我得到了响应。

感谢所有人回复我，你的回复也帮助我学习东西：）

调用.net webservice时出现问题？

2 个答案: