System.Net.Sockets.SocketException

[Can Dolgun]

Hi,

I got an exception from connect method. Demo account already connected to nj4x terminal. When I run connect method, throw an exception:

========================

System.Net.Sockets.SocketException

  HResult=0x80004005

  Message=Only one usage of each socket address (protocol/network address/port) is normally permitted

  Source=System.Net.Sockets

  StackTrace:

   at System.Net.Sockets.Socket.UpdateStatusAfterSocketErrorAndThrowException(SocketError error, String callerName)

   at System.Net.Sockets.Socket.DoBind(EndPoint endPointSnapshot, SocketAddress socketAddress)

   at System.Net.Sockets.Socket.Bind(EndPoint localEP)

   at System.Net.Sockets.TcpListener.Start(Int32 backlog)

   at nj4x.Net.NJ4XServer.<.ctor>b__10_0()

   at System.Threading.Thread.ThreadMain_ThreadStart()

   at System.Threading.ExecutionContext.RunInternal(ExecutionContext executionContext, ContextCallback callback, Object state)

   at System.Runtime.ExceptionServices.ExceptionDispatchInfo.Throw()

========================

Terminal image:

https://ibb.co/KWRDSjX

========================

C# Code:

(Note: exception throw as Undhandled Exception) I got this error for the first time

try

{

Nj4xClient.Connect("127.0.0.1", 7788, new Broker(Broker), Account, Password, true);

}

catch (MT4Exception ex)

{

return new ProcessResult(ex);

}

catch (Exception ex)

{

return new ProcessResult(ex);

}

[Roman Gerasimenko]

You need to setup different 'nj4x_server_port' values for different applications running on the same machine.

NJ4X client (Java or C#) application  automatically starts socket server, which is bound to 127.0.0.1 IP and 7777 port by default.

 

This is done for security reason, and you can change IP address and/or port using "nj4x_server_host"/"nj4x_server_port" System properties (Java) or  App.config properties (C#)

 

Each instance of application creates socket server at client machine.

It uses port number specified by nj4x_server_port App.config parameter.

 

Therefore, it is not possible to start two different applications using the same nj4x_server_port parameter.

 

P.S.

In distributed environment MT4 terminals would be able to connect to your Java/.Net application, if you set  "nj4x_server_host" system property to the available IP address of your app machine, accessible from MT4 terminal's box.

 

e.g.,

Terminal Server machine: 192.168.0.1

Java/.Net app machine: 192.168.0.123

=> C# App.config:

<configuration>...<appSettings>...

<add key="nj4x_server_host" value="192.168.0.123" />

...</appSettings>...</configuration>

...

        // strategy connects to Terminal Server as usual

            mt4.Connect( "192.168.0.1", 7788,  new Broker(p.Broker), p.Account, p.Password);

 

=> Java app class: ... static { System.setProperty("nj4x_server_host", "192.168.0.123");  } ...

        // strategy connects to Terminal Server as usual

                   strategy.connect("192.168.0.1", 7788, new Broker("..."), ...

 

 

P.S.

nj4x_server_port - this is really confusing property name, but this is not a terminal server port. It is port of your application, mt4 terminals connects to.

So, you need to set different nj4x_server_port values for different applications, e.g. 5555 for app A and 5566 for application B.

Port of TS remains unchanged (i.e. 7788) and you pass this port in the connect method: session.Connect(tsHost, tsPort=7788, ...)

нд, 12 трав. 2019 о 15:10 Can Dolgun <hc.d...@gmail.com> пише:

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