除了@galethil 的回答,如果用户打开多个选项卡(套接字连接),每个选项卡(套接字连接) 对单个用户都有唯一的套接字ID 怎么办? ,所以我们需要管理特定用户的套接字 id 数组,
客户端连接:
支持Socket IO Client v3.x,
<!-- SOCKET LIBRARY IN HTML -->
<script src="https://cdnjs.cloudflare.com/ajax/libs/socket.io/3.0.5/socket.io.js"></script>
const host = "http://yourdomain.com";
// PASS your query parameters
const queryParams = { userId: 123 };
const socket = io(host, {
path: "/pathToConnection",
transports: ['websocket'], // https://stackoverflow.com/a/52180905/8987128
upgrade: false,
query: queryParams,
reconnection: false,
rejectUnauthorized: false
});
socket.once("connect", () => {
// USER IS ONLINE
socket.on("online", (userId) => {
console.log(userId, "Is Online!"); // update online status
});
// USER IS OFFLINE
socket.on("offline", (userId) => {
console.log(userId, "Is Offline!"); // update offline status
});
});
服务器端连接:支持Socket IO Server v3.x,
const _ = require("lodash");
const express = require('express');
const app = express();
const port = 3000; // define your port
const server = app.listen(port, () => {
console.log(`We are Listening on port ${port}...`);
});
const io = require('socket.io')(server, {
path: "/pathToConnection"
});
let users = {};
io.on('connection', (socket) => {
let userId = socket.handshake.query.userId; // GET USER ID
// CHECK IS USER EXHIST
if (!users[userId]) users[userId] = [];
// PUSH SOCKET ID FOR PARTICULAR USER ID
users[userId].push(socket.id);
// USER IS ONLINE BROAD CAST TO ALL CONNECTED USERS
io.sockets.emit("online", userId);
// DISCONNECT EVENT
socket.on('disconnect', (reason) => {
// REMOVE FROM SOCKET USERS
_.remove(users[userId], (u) => u === socket.id);
if (users[userId].length === 0) {
// ISER IS OFFLINE BROAD CAST TO ALL CONNECTED USERS
io.sockets.emit("offline", userId);
// REMOVE OBJECT
delete users[userId];
}
socket.disconnect(); // DISCONNECT SOCKET
});
});
GitHub Demo