无法从Pocket PC模拟器连接到Web服务 [英] unable to connect to web service from pocket pc emulator

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问题描述


我正在尝试通过Pocket PC Phone Emulator调用在本地主机中运行的Web服务,并将其拥抱到活动同步中

以下是我的Web服务的网址
http://localhost:1043/VLR/Service.asmx

在设备程序中,我写了

Hi,
I am trying to call a web service running in localhost through pocket PC phone Emulator.I cradled it to active sync

The following is URL of my web service
http://localhost:1043/VLR/Service.asmx

In device program i wrote

private void Form1_Load(object sender, EventArgs e)
       {
           localhost.Service ls = new localhost.Service();
           string a = ls.authenticate("adssad","fssfd", "dsada");
           MessageBox.Show(a);
       }


身份验证是我在Web服务中定义的方法.
以下是我执行时的错误


Where authenticate is a method i defined in my web service.
The following are the error when i execute

System.Net.WebException was unhandled
  Message="Could not establish connection to network."
  StackTrace:
    at System.Net.HttpWebRequest.finishGetResponse()
    at System.Net.HttpWebRequest.GetResponse()
    at System.Web.Services.Protocols.WebClientProtocol.GetWebResponse()
    at System.Web.Services.Protocols.HttpWebClientProtocol.GetWebResponse()
    at System.Web.Services.Protocols.SoapHttpClientProtocol.doInvoke()
    at System.Web.Services.Protocols.SoapHttpClientProtocol.Invoke()
    at DeviceApplication2.localhost.Service.authenticate()
    at DeviceApplication2.Form1.Form1_Load()
    at System.Windows.Forms.Form.OnLoad()
    at System.Windows.Forms.Form._SetVisibleNotify()
    at System.Windows.Forms.Control.set_Visible()
    at System.Windows.Forms.Application.Run()
    at DeviceApplication2.Program.Main()



我用127.0.0.1替换了localhost.如果我输入原始IP地址,则无法加载Web服务.有什么办法吗?



I replaced localhost with 127.0.0.1. If i put my original IP address, the web service could not be loaded. Is there any way out ?

推荐答案

模拟器伪装成一个单独的设备. Localhost(127.0.0.1)将仅在您的主机上工作.尝试改用您计算机的IP地址.
The emulator pretends to be a separate device. Localhost (127.0.0.1) will only work on your host computer. Try using the IP address of your computer instead.


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