PHP - 使用JAXL登录到Facebook聊天中的问题

我尝试使用新的JAXL 3.0连接到Facebook的聊天。他们已经在他们的实例名为“echo_facebook_client.php”的例子：PHP - 使用JAXL登录到Facebook聊天中的问题

<?php 
/** 
* Jaxl (Jabber XMPP Library) 
* 
* Copyright (c) 2009-2012, Abhinav Singh <[email protected]>. 
* 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"; 

?>

我改变了JID值与Facebook账户的用户ID我尝试聊天，fb_app_key到我的应用程序和fb_access_token的应用程序ID到我请求xmpp_login特权时获得的访问令牌。

然而，当我运行它不断告诉我剧本“得到on_auth_failure CB与理由不授权”

但是没有被授权，为什么不autorized，我有XMPP权限。

然后我尝试了他们在聊天api的文档页面上的facebook示例代码，您可以在这里找到developers.facebook.com/docs/chat/。

我给了它相同的凭据，它能够在第一次尝试登录，所以错误必须在jaxl方的某处。

我对XMPP协议没有任何线索，所以使用JAXL而不是编写我自己需要的函数就好像它是在facebook示例中完成的那样，这将非常方便。

有没有人有建议，为什么JAXL代码不运行？最近的更新是8天前，所以理论上它应该工作。

来源

2012-07-20 Matthias Dunnowa