Question

我尝试使用新的JAXL 3.0连接到Facebook的聊天室。他们的例子中有一个名为“echo_facebook_client.php”的例子：

<?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值更改为我尝试与之聊天的facebook帐户的用户ID，fb_app_key更改为我的应用程序的应用程序ID，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“得到on_auth_failure cb，原因未经授权”

但是未授权的内容以及为什么不进行自动化，我有xmpp权限。

然后我在他们的文档页面上尝试了facebook的示例代码，用于聊天api，你可以在这里找到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端的某处。

我对XMPP协议一无所知，因此使用JAXL代替编写我自己需要的函数非常方便，就像在facebook示例中完成一样。

是否有人建议为什么JAXL代码不运行？最后一次更新是在8天前，所以理论上它应该可以工作。

Answer 1

您只需要将'force_tls' => true作为JAXL构造函数配置选项传递。按照https://developers.facebook.com/docs/chat/#jabber进行协商 x-facebook-platform认证必须使用。

来自jaxl groups / forums https://groups.google.com/forum/?hl=en_US&fromgroups#!topic/jaxl/Z7z32lMqG44的相关主题讨论了这个用例。

PHP - 使用JAXL登录Facebook聊天时出现的问题

1 个答案: