Remoting 如何注册多个类型？

Question

假设以下情景（Remoting）：有3个IChannel（8001,8002,8003）那么我现在要在：
端口8001和8002 注册远程类型 Type1
在8003 注册类型 Type2请问要如何实现？

Answer 1

RemotingConfiguration.RegisterWellKnownServiceType(typeof(Class.IO), "Class", WellKnownObjectMode.SingleCall);
这样不能注册多个么？

Answer 2

没办法用WCF。要求 2.0

Answer 3

除非做成 WEB服务。可是这不是所愿。

Answer 4

你不是已经指定了么？
TcpServerChannel channel = new TcpServerChannel(端口);

Answer 5

BinaryServerFormatterSinkProvider provider = new BinaryServerFormatterSinkProvider();
            provider.TypeFilterLevel = System.Runtime.Serialization.Formatters.TypeFilterLevel.Full;
            BinaryClientFormatterSinkProvider clientProvider = new BinaryClientFormatterSinkProvider();
            IDictionary props = new Hashtable();
            props["port"] = 9999;
            props["name"] = "Class1";
            TcpChannel channel = new TcpChannel(props,clientProvider,provider);
            ChannelServices.RegisterChannel(channel, false);
            RemotingConfiguration.RegisterWellKnownServiceType(typeof(自定义类), "Class1", WellKnownObjectMode.SingleCall);
            RemotingConfiguration.CustomErrorsMode = CustomErrorsModes.Off;

Answer 6

受鄙视了……囧。
这样？
ChannelServices.RegisterChannel(channel8001, false);
RemotingConfiguration.RegisterWellKnownServiceType(typeof(Type1), "Type1", WellKnownObjectMode.SingleCall);
ChannelServices.RegisterChannel(channel8002, false);
RemotingConfiguration.RegisterWellKnownServiceType(typeof(Type2), "Type2", WellKnownObjectMode.SingleCall);ChannelServices.RegisterChannel(channel8001, false);
RemotingConfiguration.RegisterWellKnownServiceType(typeof(Type3), "Type3", WellKnownObjectMode.SingleCall);

Answer 7

在配置文件里配啊,有几个配几个。例子：<system.runtime.remoting>
  <application>
    <service>
      <wellknown
        type="XXXSVC.AAAService, XXXSVC" objectUri="AAAServices" mode="SingleCall" />
      <wellknown
        type="XXXSVC.BBBService, XXXSVC" objectUri="BBBService" mode="SingleCall" />
    </service>
    <channels>
      <channel ref="tcp" port="20002">
        <serverProviders>
           <formatter ref="binary" typeFilterLevel="Full" />
        </serverProviders>
      </channel>
    </channels>
  </application>
</system.runtime.remoting>

Answer 8

你这样是不是：
20002端口可以访问 XXXSVC.AAAService和 XXXSVC.BBBService？

Answer 9

       TestClass<IDemoClass1>("tcp://localhost:8123/Demo1", c =>
                {
                    this.ShowInfo(c.Add(1, 1).ToString());
                });
            TestClass<IDemoClass1>("tcp://localhost:8124/Demo2", c =>
            {
                this.ShowInfo(c.Add(1, 2).ToString());
            });
            TestClass<IDemoClass2>("tcp://localhost:8125/Demo3", c =>
            {
                this.ShowInfo(c.Subtract(1, 3).ToString());
            });
            TestClass<IDemoClass2>("tcp://localhost:8123/Demo3", c =>
            {
                this.ShowInfo(c.Subtract(1, 5).ToString());
            });这段连接代码是完全连接成功的。
但是在 IDemoClass2 我只希望在 8125 端口上使用。

Answer 10

能做是能做，比如直接Marshal也可以实现。编码应该可以做到，我想想配置怎么搞

Answer 11

先贴代码控制的：BinaryClientFormatterSinkProvider cbin = new BinaryClientFormatterSinkProvider();
BinaryServerFormatterSinkProvider sbin = new BinaryServerFormatterSinkProvider();
sbin.TypeFilterLevel = TypeFilterLevel.Full;Hashtable properties1 = new Hashtable();
properties1["name"] = "channel1";
properties1["port"] = 8001;
TcpChannel channel1 = new TcpChannel(properties1, cbin, sbin);
ChannelServices.RegisterChannel(channel1, false);
RemotingConfiguration.RegisterWellKnownServiceType(typeof(AService), "AService", WellKnownObjectMode.SingleCall);Hashtable properties2 = new Hashtable();
properties2["name"] = "channel2";
properties2["port"] = 8002;
TcpChannel channel2 = new TcpChannel(properties2, cbin, sbin);
ChannelServices.RegisterChannel(channel2, false);
RemotingConfiguration.RegisterWellKnownServiceType(typeof(BService), "BService", WellKnownObjectMode.SingleCall);

Answer 12

那应该用 WCF 了，找了下没找到解决办法。http://stackoverflow.com/questions/234067/how-to-bind-services-to-specific-channels-tcp-http-ipc-in-a-net-remoting-s

Answer 13

上一个项目就用WCF。已经完美实现效果。但是这次强制要求2.0。
我在想想办法吧。
谢谢你。

Answer 14

利用 AppDomain ！？