Question

我使用xmpppy http://xmpppy.sourceforge.net/发送Jabber通知，使用以下代码可以很好地处理单个目标：

# pip install https://github.com/rochacbruno/xmpppy/tarball/master

import xmpp

JABBER_SETTINGS = {"USERNAME": None, "PASSWORD": None, "DOMAIN": None, "RESOURCE": None}    

def get_jabber_client():
    client = xmpp.Client(JABBER_SETTINGS.get('DOMAIN'))
    client.connect(server=(JABBER_SETTINGS.get('DOMAIN'), '5222'))
    client.auth(
        JABBER_SETTINGS.get('USERNAME'),
        JABBER_SETTINGS.get('PASSWORD'), 
        JABBER_SETTINGS.get('RESOURCE')
    )
    client.sendInitPresence()
    return client

def send_message(to, message):
    client = get_jabber_client()
    xmpp_message = xmpp.Message(to, message)
    client.send(xmpp_message)
    client.disconnect()

send_message("single.destination@domain.com", "Hello World!")

但现在我需要将消息发送到多个目的地，现在我正在做。

for users in list_of_users:
    send_message(user, "Hello World!")

哪种方法很好，但每次调用它都会启动身份验证过程并花费大量时间。

我尝试创建单个客户端并使用同一客户端发送消息。

def send_message(to, message):
    if isinstance(to, basestring):
        to = [to]
    assert isinstance(to, (list, tuple))
    client = get_jabber_client()
    for destination in to:
        xmpp_message = xmpp.Message(destination, message)
        client.send(xmpp_message)
    client.disconnect()

send_message(['user1...', 'user2...'], "Hello World!")

上面的代码有效，但只有列表中的第一个用户才能正确格式化消息，其他用户以纯XML格式接收消息。

我看到了这段代码（在.net中），提到了XEP-0033：Extended Stanza Addressing http://forum.ag-software.net/thread/1482-Send-Message-To-all-users-in-contact-list

var addresses = new Addresses();
addresses.AddAddress(new Address
              {
                  Type = Type.to,
                  Jid = "hildjj@jabber.org/Work",
                  Description = "Joe Hildebrand"
              });

addresses.AddAddress(new Address
        {
            Type = Type.cc,
            Jid = "jer@jabber.org/Home",
            Description = "Jeremie Miller"
        });

var msg = new Matrix.Xmpp.Client.Message();

msg.Add(addresses);
msg.To = "multicast.jabber.org";
msg.Body = "Hello, world!";

builds the following Xml:

<message to='multicast.jabber.org'>
   <addresses xmlns='http://jabber.org/protocol/address'>
       <address type='to' jid='hildjj@jabber.org/Work' desc='Joe Hildebrand'/>
       <address type='cc' jid='jer@jabber.org/Home' desc='Jeremie Miller'/>
   </addresses>
   <body>Hello, world!</body>
</message>

但我没有找到使用xmpppy在Python中做同样的方法，有关如何构建多播节并使用Python将消息发送给多个用户的任何想法？

感谢。

Answer 1

你可以在没有XEP-0033的情况下完成这项工作。在此循环中，您第一次覆盖变量message的值，后续消息将出现乱码内容：

for destination in to:
    message = xmpp.Message(destination, message)
    client.send(message)

请改为尝试：

for destination in to:
    xmpp_message = xmpp.Message(destination, message)
    client.send(xmpp_message)

如何使用xmpp和Python xmpppy发送多播消息（多个用户）（XEP-0033：扩展Stanza寻址）

1 个答案: