Srikanth Jeeva's technical Blog

If this is the project structure, /home/directory/lib -- Libary2.pm /home/directory/project -- file1.pl -- file2.pl -- Libary1.pm In `file1.pl`, we can include both libraries by, use lib '/home/directory/project'; use lib '/home/directory/lib'; or So in `file1.pl`, #include both directory at once, use lib qw( /home/directory/project /home/directory/lib ); use Libary1; use Libary2; `file2.pl`,will have the same use lib qw( /home/directory/project /home/directory/lib ); use Libary1; use Libary2; The above code is good for 1 or 2 files. But if there are 10+ perl files, we have to include the same lib path in all files. If the library path changes, we had to change all files with the correct path name. So what I did was to have 1 module `config/LibPaths.pm` that will have the paths, and all files will include the module. lib -- Libary2.pm project -- file1.pl -- file2.pl -- Libary1.pm config -- LibPaths.pm ...

Create a room with conversation_id and make users to subscribe to that room, so that you can emit a private message to that room it by, client ------ var socket = io.connect( 'http://ip:port' ); socket.emit( 'subscribe' , conversation_id); socket.emit( 'send message' , { room : conversation_id, message : "Some message" }); socket.on( 'conversation private post' , function (data) { //display data.message }); Server ------- socket.on( 'subscribe' , function (room) { console.log( 'joining room' , room); socket.join(room); }); socket.on( 'send message' , function (data) { console.log( 'sending room post' , data.room); socket.broadcast.to(data.room).emit( 'conversation private post' , { message : data.message }); }); Here is my Stackoverflow answer : http : //stackoverflow.com/a/23623724/453486

Took few minutes to figure this out