javascript - 维护套接字监听器的标准方法

Question

我有一个 node.js socket.io 应用程序，其中有一些不同的事件和监听器。现在我就是这样做的。

class testEmitterClass extends events {
}

const testEmitter = new testEmitterClass();

io.on('connection', function (socket) {

    console.log('connected');

    let dnsInactiveTermsListener = function (dnsInactiveTerms) {
        socket.emit(socketEvents.DNS_INACTIVE_TERMS, dnsInactiveTerms);
    };

    let checkpointInactiveTermsListener = function(checkpointInactiveTerms) {
        socket.emit(socketEvents.CHECKPOINT_INACTIVE_TERMS, checkpointInactiveTerms);
    };

    let dnsActiveTermsListener = function (dnsActiveTerms) {
        socket.emit(socketEvents.DNS_ACTIVE_TERMS, dnsActiveTerms);
    };

    let checkpointActiveTermsListener = function(checkpointActiveTerms) {
        socket.emit(socketEvents.CHECKPOINT_ACTIVE_TERMS, checkpointActiveTerms);
    };

    let dnsCountListener = function (dnsCountStreaming) {
        socket.emit(socketEvents.DNS_COUNT, dnsCountStreaming);
    };

    testEmitter.on(socketEvents.CHECKPOINT_ACTIVE_TERMS, checkpointActiveTermsListener);
    testEmitter.on(socketEvents.DNS_INACTIVE_TERMS, dnsInactiveTermsListener);
    testEmitter.on(socketEvents.CHECKPOINT_INACTIVE_TERMS, checkpointInactiveTermsListener);
    testEmitter.on(socketEvents.DNS_ACTIVE_TERMS, dnsActiveTermsListener);
    testEmitter.on(socketEvents.DNS_COUNT, dnsCountListener);

    socket.on('disconnect', function () {

        console.log('disconnected');

        testEmitter.removeListener(socketEvents.DNS_INACTIVE_TERMS, dnsInactiveTermsListener);
        testEmitter.removeListener(socketEvents.DNS_ACTIVE_TERMS, dnsActiveTermsListener);
        testEmitter.removeListener(socketEvents.DNS_COUNT, dnsCountListener);
        testEmitter.removeListener(socketEvents.CHECKPOINT_INACTIVE_TERMS, checkpointInactiveTermsListener);
        testEmitter.removeListener(socketEvents.CHECKPOINT_ACTIVE_TERMS, checkpointActiveTermsListener);

    })

});

测试发射器是一个单一实例，它在其他地方发出事件并使用 socket.io 发送到客户端

有没有办法在某处维护单个监听器列表，以便更好地维护此代码？如何将事件映射到监听器，以便在客户端与 socket.io 断开连接时添加和删除事件，而不会造成困惑。

socketEvents 只是事件名称的对象。

const DNS_COUNT = 'dnsCount';
const DNS_INACTIVE_TERMS = 'dnsInactiveTerms';
const DNS_ACTIVE_TERMS = 'dnsActiveTerms';
const CHECKPOINT_INACTIVE_TERMS = 'checkpointInactiveTerms';
const CHECKPOINT_ACTIVE_TERMS = 'checkpointActiveTerms';

module.exports = {
    DNS_COUNT,
    DNS_INACTIVE_TERMS,
    CHECKPOINT_INACTIVE_TERMS,
    DNS_ACTIVE_TERMS,
    CHECKPOINT_ACTIVE_TERMS
};

希望我说清楚了，谢谢!

Answer 1

我认为你可以改变你做事的整个方式。您可以将消息广播到所有连接的套接字，而不是为每个连接的套接字注册事件处理程序。因此，我认为您可以用以下内容替换您显示的所有内容:

class testEmitterClass extends events {
}

const testEmitter = new testEmitterClass();

const notifications = [
    CHECKPOINT_ACTIVE_TERMS,
    DNS_INACTIVE_TERMS,
    CHECKPOINT_INACTIVE_TERMS,
    CHECKPOINT_INACTIVE_TERMS,
    DNS_COUNT
];

for (let msg of notifications) {
    testEmitter.on(socketEvents[msg], function(data) {
        // send this message and data to all currently connected sockets
        io.emit(socketEvents[msg], data);
    });
}

另请注意，代码为 DRYed通过使用可以循环的消息表，而不是一遍又一遍地重复相同的语句。因此，现在要添加、删除或编辑一条通知消息，您只需在一处修改表格即可。

<小时/>

如果 socketEvents (您没有显示)只是一个具有这 5 个属性的对象，那么您甚至可以通过迭代来删除 notifications 数组socketEvents 的属性。

这将进一步减少代码:

class testEmitterClass extends events {
}

const testEmitter = new testEmitterClass();

for (let msg of Object.keys(socketEvents)) {
    testEmitter.on(socketEvents[msg], function(data) {
        // send this message and data to all currently connected sockets
        io.emit(socketEvents[msg], data);
    });
}