Question

我正在使用node-xmpp模块连接到XMPP服务器并加入群聊。到目前为止，连接到服务器，设置状态，加入房间和读取消息都有效。但我也想收到房间的用户名单。

XMPP协议要求在客户端进入会议室（http://xmpp.org/extensions/xep-0045.html#enter-pres）时发送在线节。但是我现在如何在节点中解析它呢？

我的代码目前看起来像这样：

var xmpp = require('node-xmpp');

// Create the XMPP Client
var cl = new xmpp.Client({
    jid: jid,
    password: password,
    reconnect: true
});

// Do things when online
cl.on('online', function() {
  util.log("We're online!");

  // Set client's presence
  cl.send(new xmpp.Element('presence', { type: 'available' }).c('show').t('chat'));
  cl.send(new xmpp.Element('presence', { to: room_jid+'/'+room_nick }).c('x', { xmlns: 'http://jabber.org/protocol/muc' }).c('history', {seconds: 1}));

  // Send keepalive
  setInterval(function() {
    cl.send(' ');
  }, 30000);




  cl.on('stanza', function(stanza) {
      // always log error stanzas
      if (stanza.attrs.type == 'error') {
        util.log('[error] ' + stanza);
        return;
      }

      // ignore everything that isn't a room message
      if (!stanza.is('message') || !stanza.attrs.type == 'chat') {
        return;
      }

      var body = stanza.getChild('body');
      // message without body is probably a topic change
      if (!body) {
        return;
      }

    // Extract username
    var from, room, _ref;
    _ref = stanza.attrs.from.split('/'), room = _ref[0], from = _ref[1];
     var message = body.getText();

     // Log topics and messages to the console
     if(!from) {
        util.log('Topic: ' + message);
     } else {
        util.log('[' + from + ']: ' + message);
     }
    });
});

我已尝试使用

触发状态

if(stanza.is('presence')) {}

在cl.on（'stanza'）部分，但它不起作用。

Answer 1

UPDATE ：我现在正在描述一种新方法，它不需要客户端发送请求。

后台：当客户端加入群聊时，服务器会返回在线节，其中包含有关已连接用户的群聊信息。

cl.on('stanza', function(stanza) {
    // always log error stanzas
    if (stanza.attrs.type == 'error') {
        util.log('[error] ' + stanza);
        return;
    }

    if(stanza.is('presence')){
        // We are only interested in stanzas with <x> in the payload or it will throw some errors
        if(stanza.getChild('x') !== undefined) {
            // Deciding what to do based on the xmlns attribute
            var _presXmlns = stanza.getChild('x').attrs.xmlns;

            switch(_presXmlns) {
                // If someone is joining or leaving
                case 'http://jabber.org/protocol/muc#user':
                    // Get the role of joiner/leaver
                    _presRole = stanza.getChild('x').getChild('item').attrs.role;
                    // Get the JID of joiner/leaver
                    _presJID  = stanza.getChild('x').getChild('item').attrs.jid;
                    // Get the nick of joiner/leaver
                    _presNick = stanza.attrs.from.split('/')[1];


                    // If it's not none, this user must be joining or changing his nick
                    if(_presRole !== 'none') {

                        // We are now handling the data of joinging / nick changing users. I recommend to use an in-memory store like 'dirty' [https://github.com/felixge/node-dirty] to store information of the users currentliy in the group chat.


                    } else {

                        // We are now handling the data of leaving users

                    }
                break;
            }

            return;
        }

        return;
    }

OLD METHOD

我之前描述过如何在群聊中为当前用户查询服务器的方法。通过维护存储所有用户流量（加入，离开，缺口更改）的存储，不再需要这样做。但是，您仍然可以使用它来确保数据是否一致，因为存在节未正确传递给客户端。这就是下面仍然描述的原因：

要请求连接到房间的用户的列表，您需要执行以下操作：

首先向服务器发送请求并询问用户列表：

cl.send(new xmpp.Element('iq', {from: jid, to: room_jid, type: 'get' }).c('query', { xmlns: 'http://jabber.org/protocol/disco#items' }));

然后听取iq-stanzas，解析它们并用数据填充数组：

// Catching the requested user list
if(stanza.is('iq')){
    // Fetching usernames from return data (data structure: http://xmpp.org/extensions/xep-0045.html#example-12)
    var _items = stanza.getChild('query').getChildren('item');
    var users = new Array();
    for(var i = 0; i<_items.length; i++) {
        // We are building an object here to add more data later
        users[i] = new Object();
        users[i]['name'] = _items[i].attrs.name;
    }
    console.log(util.inspect(users, {depth: null, colors: true}));
    return;
}

这将为您提供用户列表。要请求唯一的JID，您必须探测每个用户。要使列表保持最新，您应该在用户离开时删除用户，并在他们加入时添加+ probe。

使用Node处理XMPP存在

1 个答案:

OLD METHOD