websocket send parameter in connection?

Question

I have a scenario which I need to create a very simple text chat using nodejs and Websocket.

I have everything set up and working.

I now need to allow users to send a parameter in the websocket connection URL.

Like so: websocket = new WebSocket("ws://test-website.net:8080/?id=55");

in my reserach I came across quite a few similar questions and found out that I can pass the parameters in the URL like the example above.

However, none of them mentions or explains how to get this parameter (55 in this example) in server.js

wss.on('connection', function(ws) {

///I need to get the parameter here////

});

Could someone please advice on this issue?

Thanks in advance.

Answer 1

In your websocket request handler, it is normally passed the web socket as the first argument; the incoming HTTP request is passed as a second argument:

webSocketServer.on('connection', function connection(ws_client_stream, incoming_request) {
  console.log(incoming_request.url);
});

From there, you can use url.parse to get the components of the URL, such as the query string that you are interested in.

Answer 2

For websocket library in nodejs (websocket.js npm) :

wsServer.on('request', function(request){
    var FullURL = request.resourceURL.query;
    console.log(FullURL);
    // . . your code  . . //
}

link = wss://localhost/?param=value
output

{ param : 'value' }

here request.resourceURL retuen a object

Url {
  protocol: null,
  slashes: null,
  auth: null,
  host: null,
  port: null,
  hostname: null,
  hash: null,
  search: '?id=65&encoding=text',
  query: { id: '65', encoding: 'text' },
  pathname: '/',
  path: '/?id=65&encoding=text',
  href: '/?id=65&encoding=text' }

Answer 3

If you're using 'websocket' lib, the following code retrieves your queryStrings:

wsServer.on('request', function(request) 
{
    console.log(request.resourceURL.query);
}

Note 1: You don't need to parse anything here.

Note 2: Though it has been answered years ago, this answer goes to those in case where people find the above-mentioned answer more complicated.