c# - 'agsXMPP.MessageGrabber' 是一个 'type' 但像 'variable' 一样使用

Question

代码如下:

  XmppClientConnection xmpp = new XmppClientConnection();
        xmpp.Server = "gmail.com";
        xmpp.ConnectServer = "talk.google.com";
        xmpp.Username = "aleksandr.gordon";
        xmpp.Password = "password";
        xmpp.Open();

        agsXMPP.Jid JID = new Jid("thegabmeister1@gmail.com");
        xmpp.MessageGrabber.Add(JID, new agsXMPP.Collections.BareJidComparer(), new MessageCB(xmpp.MessageGrabber), null);

        agsXMPP.protocol.client.Message msg = new agsXMPP.protocol.client.Message();
        msg.Type = agsXMPP.protocol.client.MessageType.chat;
        msg.To = JID;
        msg.Body = "how u doing" + DateTime.Now.ToString();

        xmpp.OnLogin += delegate(object o) { xmpp.Send(msg); };

更新的新代码:

    agsXMPP.Jid JID = new Jid("thegabmeister@gmail.com");
    xmpp.MessageGrabber.Add(JID, new agsXMPP.Collections.BareJidComparer(),
        new MessageCB(delegate(object sender, agsXMPP.protocol.client.Message msg, object data)
    {
    }), null);
    agsXMPP.protocol.client.Message msg1 = new agsXMPP.protocol.client.Message();
    msg1.Type = agsXMPP.protocol.client.MessageType.chat;
    msg1.To = JID;
    msg1.Body = "how u doing" + DateTime.Now.ToString();

    xmpp.OnLogin += delegate(object o) { xmpp.Send(msg1); };

我做错了什么？为什么会出现此错误？

Answer 1

您可能需要传递一个委托(delegate)方法，尝试将此方法添加到您的代码中

void Login() {
    XmppClientConnection xmpp = new XmppClientConnection();
    xmpp.Server = "gmail.com";
    xmpp.ConnectServer = "talk.google.com";
    xmpp.Username = "aleksandr.gordon";
    xmpp.Password = "password";
    xmpp.Open();

    agsXMPP.Jid JID = new Jid("thegabmeister1@gmail.com");
    xmpp.MessageGrabber.Add(JID, new agsXMPP.Collections.BareJidComparer(), new MessageCB(GrabMessage), null);

    agsXMPP.protocol.client.Message msg = new agsXMPP.protocol.client.Message();
    msg.Type = agsXMPP.protocol.client.MessageType.chat;
    msg.To = JID;
    msg.Body = "how u doing" + DateTime.Now.ToString();

    xmpp.OnLogin += delegate(object o) { xmpp.Send(msg); };
}

protected void GrabMessage(object sender, agsXMPP.protocol.client.Message msg, object data) {
}

这是我能建议的最后一件事，但上述方法有效并将解决您的问题，除非您在其他地方有其他问题:

xmpp.MessageGrabber.Add(JID, new agsXMPP.Collections.BareJidComparer(),
    new MessageCB(delegate(object sender, agsXMPP.protocol.client.Message msg, object data) {
    if (msg.Body != null) {
        MessageBox.Show(msg.Body);
    }
    }), null);