[英]Can anyone tell me what's wrong with this switch statement (javascript)?
[英]Can anyone please tell me what's wrong this socket event?
我在用户加入和离开时发出了两个事件( user_joined
和user_left
)。 它在服务器端工作,但不在客户端工作。
服务器端代码:(它正在工作,在每个连接上显示 console.log)
io.on('connection', function (socket) {
const id = socket.id;
/**
* User Join Function
*/
socket.on('join', function ({ name, room }) {
const { user } = addUser({id, name, room}); // add user to users array
socket.join(user.room);
socket.emit('user_joined', users); // emit event with modified users array
console.log(id, 'joined')
})
/**
* User Disconnect function
*/
socket.on('disconnect', () => {
removeUser(id); // remove user form users array
socket.emit('user_left', users); // emit event with modified users array
console.log(id, 'left')
})
})
客户端代码:(未在user_joined
或user_left
上触发)
const [players, setPlayers] = useState([]);
const ENDPOINT = 'localhost:5000';
socket = io(ENDPOINT);
useEffect(() => {
const name = faker.name.firstName() + ' ' + faker.name.lastName();
socket.emit('join', {name, room: 'global'}); // it's working fine
return () => {
socket.emit('disconnect');
socket.off();
}
}, [])
useEffect(() => {
socket.on('user_joined', (users) => {
setPlayers(users);
}); // >>> Not Working <<<
socket.on('user_left', (users) => {
setPlayers(users);
}); // >>> Not Working <<<
console.log(socket) // it's working fine
}, [players]);
socket
实例只需要创建一次。 在您的情况下,它会在每次重新渲染时创建。 您也不需要 2 个useEffects
。
将socket
实例的创建并将您的 2 个useEffects
合并为 1 并提供一个空数组作为依赖项。 这样,您的 useEffect 只执行一次,而不是在每次重新渲染时执行。
尝试这个
const [players, setPlayers] = useState([]);
useEffect(() => {
const ENDPOINT = 'localhost:5000';
socket = io(ENDPOINT);
const name = faker.name.firstName() + ' ' + faker.name.lastName();
socket.emit('join', {name, room: 'global'});
socket.on('user_joined', (users) => {
setPlayers(users);
});
socket.on('user_left', (users) => {
setPlayers(users);
});
console.log(socket);
return () => {
socket.emit('disconnect');
socket.off();
}
}, []);
...
如果您想在组件的其他位置使用套接字实例,请使用useRef
。 使用 useRef,除非您对其进行变异,否则您始终会获得相同的实例。
使用refs
创建套接字
...
const [players, setPlayers] = useState([]);
const ENDPOINT = 'localhost:5000';
const socketInstance = useRef(io(ENDPOINT));// in react, with useRef, you always get the same instance unless you mutate it.
useEffect(() => {
// socketInstance.current = io(ENDPOINT);
const name = faker.name.firstName() + ' ' + faker.name.lastName();
socketInstance.current.emit('join', {name, room: 'global'});
socketInstance.current.on('user_joined', (users) => {
setPlayers(users);
});
socketInstance.current.on('user_left', (users) => {
setPlayers(users);
});
console.log(socketInstance.current);
return () => {
socketInstance.current.emit('disconnect');
socketInstance.current.off();
}
}, []);
...
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