Socket.IO处理断开连接事件
Socket.IO处理断开连接事件
提问于 2013-06-25 08:33:02
无法处理此disconnect事件，不知道为什么套接字未发送到客户端/客户端没有响应！
服务器
io.sockets.on('connection', function (socket) {
socket.on('NewPlayer', function(data1) {
online = online + 1;
console.log('Online players : ' + online);
console.log('New player connected : ' + data1);
Players[data1] = data1;
console.log(Players);
});
socket.on('DelPlayer', function(data) {
delete Players[data];
console.log(Players);
console.log('Adios' + data);
});
socket.on('disconnect', function () {
socket.emit('disconnected');
online = online - 1;
});
});客户端
var socket = io.connect('http://localhost');
socket.on('connect', function () {
person_name = prompt("Welcome. Please enter your name");
socket.emit('NewPlayer', person_name);
socket.on('disconnected', function() {
socket.emit('DelPlayer', person_name);
});
});正如您所看到的，当客户端断开连接时，应该删除阵列objectperson_name，但事实并非如此。
回答 3
Stack Overflow用户
回答已采纳
发布于 2013-06-26 12:41:07
好的，不是通过名字来识别玩家，而是跟踪他们所连接的套接字。你可以有这样的实现
服务器
var allClients = [];
io.sockets.on('connection', function(socket) {
allClients.push(socket);
socket.on('disconnect', function() {
console.log('Got disconnect!');
var i = allClients.indexOf(socket);
allClients.splice(i, 1);
});
});希望这能帮助你以另一种方式思考
Stack Overflow用户
发布于 2018-04-25 08:47:19
创建一个Map或Set，并使用"on connection“事件设置每个已连接的套接字，反过来，"once disconnect”事件将该套接字从我们之前创建的Map中删除
import * as Server from 'socket.io';
const io = Server();
io.listen(3000);
const connections = new Set();
io.on('connection', function (s) {
connections.add(s);
s.once('disconnect', function () {
connections.delete(s);
});
});Stack Overflow用户
发布于 2019-12-22 22:56:56
如果你愿意，你也可以像这样使用socket id来管理你的播放器列表。
io.on('connection', function(socket){
socket.on('disconnect', function() {
console.log("disconnect")
for(var i = 0; i < onlineplayers.length; i++ ){
if(onlineplayers[i].socket === socket.id){
console.log(onlineplayers[i].code + " just disconnected")
onlineplayers.splice(i, 1)
}
}
io.emit('players', onlineplayers)
})
socket.on('lobby_join', function(player) {
if(player.available === false) return
var exists = false
for(var i = 0; i < onlineplayers.length; i++ ){
if(onlineplayers[i].code === player.code){
exists = true
}
}
if(exists === false){
onlineplayers.push({
code: player.code,
socket:socket.id
})
}
io.emit('players', onlineplayers)
})
socket.on('lobby_leave', function(player) {
var exists = false
for(var i = 0; i < onlineplayers.length; i++ ){
if(onlineplayers[i].code === player.code){
onlineplayers.splice(i, 1)
}
}
io.emit('players', onlineplayers)
})
})页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/17287330
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