C# 如何在远程处理中使用singleton和singleCall

我已经创建了一个远程处理应用程序。Librery类中的代码是

 public class InitialClass1:MarshalByRefObject
{
    int i=0;
    public InitialClass1()
    {

    }
    public string getInitial(string nam)
    {                
        i++;
        if (nam.ToLower() == "naresh") return i.ToString()+").Jadapalli";
        else if (nam.ToLower() == "balu") return i.ToString() + "Gonugunta";
        else if (nam.ToLower() == "murali") return i.ToString() + "Vempuluri";
        else if (nam.ToLower() == "chandra sekhar") return i.ToString() + "Ponnam";
        else if (nam.ToLower() == "aneev") return i.ToString() + "Katti";
        else if (nam.ToLower() == "rajini") return i.ToString() + "Karlapudi";
        else
            return i.ToString() + "No results";

    }
}

控制台应用程序中的代码是

class Program
{
    [STAThread]
    static void Main(string[] args)
    {
        TcpChannel tcp = new TcpChannel(1234);
        ChannelServices.RegisterChannel(tcp,false);
        string s = ConfigurationManager.AppSettings["remote"];
        RemotingConfiguration.RegisterWellKnownServiceType(typeof(InitialClass1),s,WellKnownObjectMode.SingleCall);
        Console.WriteLine("Remoting starting...");
        Console.ReadLine();            
    }
}

 private void button1_Click(object sender, EventArgs e)
    {
        label3.Text = Icls.getInitial(textBox1.Text);
    }

    private void Form1_Load(object sender, EventArgs e)
    {

         string url = ConfigurationManager.AppSettings["remote"];
        TcpChannel chan = new TcpChannel();
        ChannelServices.RegisterChannel(chan, false);

        RemotingConfiguration.ApplicationName = "TestRemoting";
        RemotingConfiguration.RegisterActivatedClientType(typeof(InitialClass1), url);

        Icls = new InitialClass1();

    }

Windows应用程序中的代码是

class Program
{
    [STAThread]
    static void Main(string[] args)
    {
        TcpChannel tcp = new TcpChannel(1234);
        ChannelServices.RegisterChannel(tcp,false);
        string s = ConfigurationManager.AppSettings["remote"];
        RemotingConfiguration.RegisterWellKnownServiceType(typeof(InitialClass1),s,WellKnownObjectMode.SingleCall);
        Console.WriteLine("Remoting starting...");
        Console.ReadLine();            
    }
}

 private void button1_Click(object sender, EventArgs e)
    {
        label3.Text = Icls.getInitial(textBox1.Text);
    }

    private void Form1_Load(object sender, EventArgs e)
    {

         string url = ConfigurationManager.AppSettings["remote"];
        TcpChannel chan = new TcpChannel();
        ChannelServices.RegisterChannel(chan, false);

        RemotingConfiguration.ApplicationName = "TestRemoting";
        RemotingConfiguration.RegisterActivatedClientType(typeof(InitialClass1), url);

        Icls = new InitialClass1();

    }

它显示异常，因为找不到请求的服务

App.Config文件中的代码为

<configuration>
  <appSettings>
    <add key="remote" value="tcp://localhost:1234/TestRemoting"/>
  </appSettings>
</configuration>

---

我认为您应该尝试button1\u click方法中的初始化。因为您希望在单击按钮时初始化它。

我的错误在于服务器对象是控制台应用程序

TcpChannel tcp = new TcpChannel(1234);
    ChannelServices.RegisterChannel(tcp,false);
    string s = ConfigurationManager.AppSettings["remote"];
    RemotingConfiguration.RegisterWellKnownServiceType(typeof(InitialClass1),s,WellKnownObjectMode.SingleCall);
    Console.WriteLine("Remoting starting...");
    Console.ReadLine(); 

请参见注册中的“我正在从app.config文件传递整个url的s值”tcp://localhost:1234/TestRemoting. 这就是问题所在。事实上，我们必须根据您的意愿在服务器对象中提供所需的名称，但在app.config文件中应该只有相同的名称

这里是我的网址tcp://localhost:1234/TestRemoting . 在这种情况下，我们需要将服务器对象中的TestRemoting设置为

RemotingConfiguration.RegisterWellKnownServiceType(typeof(InitialClass1),"TestRemoting",WellKnownObjectMode.SingleCall);

---

那只是我的问题。现在我的应用程序运行良好

谢谢你的回答。如果我这么做的话。。i值将被初始化事件它是单音：否。。我在客户端对代码做了一些修改。现在显示请求的服务未找到错误。。