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我正在为 android 中的 DBConnection 起诉 asp.net webservice (visual studio 2008)。

它运行成功。

我的安卓代码是：

protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        final  AlertDialog ad=new AlertDialog.Builder(this).create();


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

        Button btnCall=(Button)findViewById(R.id.btnCall);

        btnCall.setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View v) {
                // TODO Auto-generated method stub

                CallSoap cs=new CallSoap();

                try
                {
                    String name=cs.Call();
                    Toast.makeText(getApplicationContext(), name, Toast.LENGTH_LONG).show();
                }
                catch(Exception ex)
                {
                    ad.setMessage(ex.getMessage());
                }
            }
        });

    }

和 callsoap 方法：

import java.util.ArrayList;

import org.ksoap2.SoapEnvelope; 
import org.ksoap2.serialization.PropertyInfo; 
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.HttpTransportSE;

import android.widget.Toast;

public class CallSoap 
{ 
    public final String SOAP_ACTION ="http://tempuri.org/GetData";


    public  final String OPERATION_NAME = "GetData"; 

    public  final String WSDL_TARGET_NAMESPACE = "http://tempuri.org/";

    public  final String SOAP_ADDRESS = "http://10.0.2.2:8080/Service1.asmx";

    public CallSoap() 
    { 
    }


    public String  Call()
    {
        SoapObject req=new SoapObject(WSDL_TARGET_NAMESPACE, OPERATION_NAME);

        SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(
                SoapEnvelope.VER11);
                envelope.dotNet = true;
                //envelope.headerOut = security; // this is an Element[] created before
                envelope.encodingStyle = SoapEnvelope.ENC;
                envelope.setAddAdornments(false);
                envelope.implicitTypes = false;
     envelope.setOutputSoapObject(req);

     HttpTransportSE  httpTransport = new HttpTransportSE(SOAP_ADDRESS);
     Object response=null;


     try
     {
     httpTransport.call(SOAP_ACTION, envelope);
     response = envelope.getResponse();
     //Toast.makeText(this, "this is my Toast message!!! =)", Toast.LENGTH_LONG).show();
     }
     catch (Exception exception)
     {
     response=exception.toString();
     }
     return response.toString();
     }

}

当我单击应用程序上的按钮时，它给了我错误：

android org.xmlpull.v1.xmlpullparserexception 预期 start_tag http //schemas.xmlsoap.org/soap/envelope/ 信封位置开始标签

我在堆栈上有很多关于这个的问题。

我发现它通常是由于编写了错误的服务方法而发生的。

但就我而言，我检查了所有这些。每件事都是正确的。

您还可以检查：我有我正在运行的服务的快照，其中给出了每一件事：

在此处输入图像描述

请帮我。

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1 回答 1

1

我认为您必须输入端口号而不是 8080

于 2013-08-21T12:24:18.333 回答