在模拟器中使用kSOAP2使用localhost WCF服务(Consuming a localhost WCF service using kSOAP2 in emulator)

 我已经把头撞到了这堵墙上了几天，所以任何帮助都非常感激。 
 
 首先是一些背景知识。 我是一位经验丰富的WindowsMo​​bile开发人员，因此我是Java，Android甚至WCF的老手。 我已经做了很多关于如何使用kSOAP2在Android中使用WCF应用程序的研究，但无论我做得最好，我能想到的是由于超时而导致Android中的套接字错误。 
 
 首先，我在Visual Studio 2008中创建了一个不需要参数的Web服务应用程序，只需使用字符串值进行响应。 Web服务代码如下： 
 
using System;
using System.Collections.Generic;
using System.Web;
using System.Web.Services;

namespace Sample
{
    [WebService(Namespace = "http://sample.com/")]
    public class Service1 : System.Web.Services.WebService
    {
        [WebMethod]
        public string SayHello()
        {
            return "Hello, Android. From your friend, WCF";
        }
    }
}
 
 对于此Web服务，我没有调整任何其他设置或对Web.config文件进行任何修改。 
 
 当我运行该服务时，它会打开我的浏览器并指向以下网址： 
 
http://localhost:61554/Service1.asmx
 
 接下来，我跳入Eclipse并创建了一个简单的Android项目来使用WCF服务。 对于初学者，我更改了AndroidManifest.xml文件并将此语句添加到其中： 
 
<uses-permission android:name="android.permission.INTERNET" />
 
 现在我的Android类的代码应该是繁重的工作： 
 
package com.sample;

import android.app.Activity;
import android.os.Bundle;
import android.widget.TextView;

import org.ksoap2.*;
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.*;

public class Sample extends Activity {
    TextView result;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

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

        try {
            SoapObject Request = new SoapObject("http://sample.com/", "SayHello");

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

            HttpTransportSE androidHttpTransport = new HttpTransportSE("http://192.168.1.72:61554/Service1.asmx");

            androidHttpTransport.call("http://192.168.1.72:61554/SayHello", envelope);
            SoapObject response = (SoapObject)envelope.getResponse();
            String resultValue =  response.getProperty(0).toString();

            result.setText(resultValue);            
        } 
        catch (Exception e) {
            result.setText(e.getMessage());
        }
    }
}
 
 给我错误的代码行是： 
 
androidHttpTransport.call("http://192.168.1.72:61554/SayHello", envelope);
 
 当在模拟器中运行时，代码到达该行，暂停一两分钟，然后进入具有超时异常的catch块。 
 
 关于我可能在这里缺少的任何想法？ 感谢您的帮助。 

I have rammed my head against this wall for a couple of days now so any help is very appreciated.
 
A bit of background, first. I am a seasoned WindowsMobile developer moving into Android so I am therefore a noob in Java, Android, and even WCF. I have done a lot of research on how to consume WCF apps in Android using kSOAP2 but no matter what I do the best I can come up with is a socket error in Android because of a time out.
 
First, I created a web service application in Visual Studio 2008 that requires no parameters and simply responds with a string value. The web service code is as follows:
 
using System;
using System.Collections.Generic;
using System.Web;
using System.Web.Services;

namespace Sample
{
    [WebService(Namespace = "http://sample.com/")]
    public class Service1 : System.Web.Services.WebService
    {
        [WebMethod]
        public string SayHello()
        {
            return "Hello, Android. From your friend, WCF";
        }
    }
}
 
For this web service I did not adjust any other settings or make any modifications to the Web.config file.
 
When I run the service it opens up my browser and points to the following url:
 
http://localhost:61554/Service1.asmx
 
Next, I jumped into Eclipse and created a simple Android project to consume the WCF service. For starters I changed the AndroidManifest.xml file and added this statement to it:
 
<uses-permission android:name="android.permission.INTERNET" />
 
Now for the code from my Android class that is supposed to do the heavy lifting:
 
package com.sample;

import android.app.Activity;
import android.os.Bundle;
import android.widget.TextView;

import org.ksoap2.*;
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.*;

public class Sample extends Activity {
    TextView result;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

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

        try {
            SoapObject Request = new SoapObject("http://sample.com/", "SayHello");

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

            HttpTransportSE androidHttpTransport = new HttpTransportSE("http://192.168.1.72:61554/Service1.asmx");

            androidHttpTransport.call("http://192.168.1.72:61554/SayHello", envelope);
            SoapObject response = (SoapObject)envelope.getResponse();
            String resultValue =  response.getProperty(0).toString();

            result.setText(resultValue);            
        } 
        catch (Exception e) {
            result.setText(e.getMessage());
        }
    }
}
 
The line of code that is giving me the error is:
 
androidHttpTransport.call("http://192.168.1.72:61554/SayHello", envelope);
 
When run in the emulator the code gets to that line, pauses for a minute or two, and then falls into the catch block with a time out exception.
 
Any ideas on what I might be missing here? The help is appreciated.

原文：https://stackoverflow.com/questions/6538055