PHP-使用JAXL登录Facebook聊天时出现的问题

我尝试使用新的JAXL 3.0连接到Facebook的聊天。他们的示例中有一个名为"echo_facebook_client.php"的示例：

<?php
/**
 * Jaxl (Jabber XMPP Library)
 *
 * Copyright (c) 2009-2012, Abhinav Singh <me@abhinavsingh.com>.
 * All rights reserved.
 *
 * Redistribution and use in source and binary forms, with or without
 * modification, are permitted provided that the following conditions
 * are met:
 *
 * * Redistributions of source code must retain the above copyright
 * notice, this list of conditions and the following disclaimer.
 *
 * * Redistributions in binary form must reproduce the above copyright
 * notice, this list of conditions and the following disclaimer in
 * the documentation and/or other materials provided with the
 * distribution.
 *
 * * Neither the name of Abhinav Singh nor the names of his
 * contributors may be used to endorse or promote products derived
 * from this software without specific prior written permission.
 *
 * THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS
 * "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT
 * LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS
 * FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE
 * COPYRIGHT OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT,
 * INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING,
 * BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES;
 * LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER
 * CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRIC
 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN
 * ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE
 * POSSIBILITY OF SUCH DAMAGE.
 *
 */
if($argc != 4) {
    echo "Usage: $argv[0] fb_user_id_or_username fb_app_key fb_access_token'n";
    exit;
}
//
// initialize JAXL object with initial config
//
require_once 'jaxl.php';
$client = new JAXL(array(
    // (required) credentials
    'jid' => $argv[1].'@chat.facebook.com',
    'fb_app_key' => $argv[2],
    'fb_access_token' => $argv[3],
    // (required) force facebook oauth
    'auth_type' => 'X-FACEBOOK-PLATFORM',
    // (optional)
    //'resource' => 'resource'
));
//
// add necessary event callbacks here
//
$client->add_cb('on_auth_success', function() {
    global $client;
    _debug("got on_auth_success cb, jid ".$client->full_jid->to_string());
    $client->set_status("available!", "dnd", 10);
});
$client->add_cb('on_auth_failure', function($reason) {
    global $client;
    $client->send_end_stream();
    _debug("got on_auth_failure cb with reason $reason");
});
$client->add_cb('on_chat_message', function($stanza) {
    global $client;
    // echo back incoming message stanza
    $stanza->to = $stanza->from;
    $stanza->from = $client->full_jid->to_string();
    $client->send($stanza);
});
$client->add_cb('on_disconnect', function() {
    _debug("got on_disconnect cb");
});
//
// finally start configured xmpp stream
//
$client->start();
echo "done'n";
?>

我将jid值更改为我尝试聊天的facebook帐户的用户id，将fb_app_key更改为我的应用程序的应用程序id，并将fb_access_token更改为我在请求xmpp_login权限时获得的访问令牌。

然而，当我运行脚本时，它一直告诉我"got on_auth_failure cb with reason not authorized"

但是什么是未授权的，为什么不授权，我有xmpp权限。

然后我在他们的文档页面上尝试了facebook的聊天api示例代码，你可以在这里找到开发者。facebook.com/docs/chat/.

我给了它相同的凭据，它能够在第一次尝试时登录，所以错误一定在jaxl端的某个地方。

我对XMPP协议一无所知，所以使用JAXL将非常方便，而不是像在facebook示例中那样自己编写所需的函数。

有人对JAXL代码为什么不运行有什么建议吗？最后一次更新是在8天前，所以理论上它应该可以工作。