How to construct a WebSocket URI relative to the page URI?

Question

I want to construct a WebSocket URI relative to the page URI at the browser side. Say, in my case convert HTTP URIs like

http://example.com:8000/path
https://example.com:8000/path

to

ws://example.com:8000/path/to/ws
wss://example.com:8000/path/to/ws

What I'm doing currently is replace the first 4 letters "http" by "ws", and append "/to/ws" to it. Is there any better way for that?

Answer 1

Here is my version which adds the tcp port in case it's not 80 or 443:

function url(s) {
    var l = window.location;
    return ((l.protocol === "https:") ? "wss://" : "ws://") + l.hostname + (((l.port != 80) && (l.port != 443)) ? ":" + l.port : "") + l.pathname + s;
}

Edit 1: Improved version as by suggestion of @kanaka :

function url(s) {
    var l = window.location;
    return ((l.protocol === "https:") ? "wss://" : "ws://") + l.host + l.pathname + s;
}

Edit 2: Nowadays I create the WebSocket this:

var s = new WebSocket(((window.location.protocol === "https:") ? "wss://" : "ws://") + window.location.host + "/ws");

Answer 2

Using the Window.URL API - https://developer.mozilla.org/en-US/docs/Web/API/Window/URL

Works with http(s), ports etc.

var url = new URL('/path/to/websocket', window.location.href);

url.protocol = url.protocol.replace('http', 'ws');

url.href // => ws://www.example.com:9999/path/to/websocket

Answer 3

Assuming your WebSocket server is listening on the same port as from which the page is being requested, I would suggest:

function createWebSocket(path) {
    var protocolPrefix = (window.location.protocol === 'https:') ? 'wss:' : 'ws:';
    return new WebSocket(protocolPrefix + '//' + location.host + path);
}

Then, for your case, call it as follows:

var socket = createWebSocket(location.pathname + '/to/ws');

Answer 4

easy:

location.href.replace(/^http/, 'ws') + '/to/ws'
// or if you hate regexp:
location.href.replace('http://', 'ws://').replace('https://', 'wss://') + '/to/ws'

Answer 5

On localhost you should consider context path.

function wsURL(path) {
    var protocol = (location.protocol === 'https:') ? 'wss://' : 'ws://';
    var url = protocol + location.host;
    if(location.hostname === 'localhost') {
        url += '/' + location.pathname.split('/')[1]; // add context path
    }
    return url + path;
}

Answer 6

In typescript:

export class WebsocketUtils {

    public static websocketUrlByPath(path) {
        return this.websocketProtocolByLocation() +
            window.location.hostname +
            this.websocketPortWithColonByLocation() +
            window.location.pathname +
            path;
    }

    private static websocketProtocolByLocation() {
        return window.location.protocol === "https:" ? "wss://" : "ws://";
    }

    private static websocketPortWithColonByLocation() {
        const defaultPort = window.location.protocol === "https:" ? "443" : "80";
        if (window.location.port !== defaultPort) {
            return ":" + window.location.port;
        } else {
            return "";
        }
    }
}

Usage:

alert(WebsocketUtils.websocketUrlByPath("/websocket"));

Answer 7

Dead easy solution, ws and port, tested:

var ws = new WebSocket("ws://" + window.location.host + ":6666");

ws.onopen = function() { ws.send( .. etc

Answer 8

I agree with @Eadz, something like this is cleaner and safer:

const url = new URL('./ws', location.href);
url.protocol = url.protocol.replace('http', 'ws');
const webSocket = new WebSocket(url);

The URL class saves work and deals with things like query parameters, etc.