Get a variable out of socket.on

Question

I have tried now to get a variable out of a socket.on(); but can't get it out as it says it is an undefined variable.

Here is the code bit that doesn't work:

 io.on('connection', function(socket){
    socket.join("0");
    socket.on("chatroom",function(chat){
        socket.join(chat);
        if(chat != "0"){
        console.log("A user joined chatroom: " + chat);
        }
        var chatroom = chat;
    });

}); 

var chatroom = chatroom;

As you see it is the variable above that get's a value of undefined even if it has a value of 2 in the io.on fuction. Any ideas how to get it out?

Excuse me for my noobish codes (only 3 days experience in node.js)

It sounds like you maybe need to read this: [How to return a value from an asynchronous operation](http://stackoverflow.com/questions/14220321/how-do-i-return-the-response-from-an-asynchronous-call/14220323#14220323). The structure of what you're trying to do is just wrong. Async results must be consumed IN the async response, not elsewhere. — jfriend00, Dec 16 '15 at 03:17

score 0 · Answer 1 · answered Dec 15 '15 at 20:32

0

I think the external chatroom variable is eclipsed by the internal one. Try:

var chatroom;
io.on('connection', function(socket) {
    socket.join("0");
    socket.on("chatroom", function(chat) {
        socket.join(chat);
        if(chat != "0") {
            console.log("A user joined chatroom: " + chat);
        }
        chatroom = chat;
    });    
});

Also, you have to understand, that because the code is asynchronous, the last line of your code will be executed before the code inside of the io.on('connection', ...)

answered Dec 15 '15 at 20:32

Filip Spiridonov

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Oh okay I understand but how do I make it execute before it executes the rest? – Oliver Weitman Dec 15 '15 at 21:04
I also got an error from that as it says the variable is undefined... Could it be the same issue as it executes the code asynchronous – Oliver Weitman Dec 15 '15 at 21:06
I'm missing a context here.. what do you want to do with that `chatroom` variable? Variable on which line is undefined? – Filip Spiridonov Dec 15 '15 at 23:58
The value of variable chatroom gets undefined outside the socket.on even if it is 4 inside it – Oliver Weitman Dec 16 '15 at 12:44
The `chatroom` variable will be defined after the `socket.on(...)` is executed. If you use it before that, it is undefined. Try inserting the `console.log(..)` statements to understand the order of execution. – Filip Spiridonov Dec 16 '15 at 22:23

Get a variable out of socket.on

1 Answers1