如何将$ _SESSION变量传递给websocket服务器?

ton*_*nix 6 php apache session websocket

我在网上搜索了很多,但没有找到有用的线索.

我有一个websocket服务器和一个在我的本地机器上一起运行的Web服务器.

当客户端使用浏览器API'new WebSocket("ws:// localhost")'连接到它时,我需要将$ _SESSION数据传递给websocket服务器(请求使用反向代理发送到websocket,后者知道它当收到带有"升级"标题的请求时.

关键是客户端成功连接到ws服务器,但我还需要使用HTTP Web服务器设置的$ _SESSION变量来恢复其SESSION数据.

其实我的情况是这样的(我正在使用Ratchet库):

use Ratchet\Server\IoServer;
use Ratchet\Http\HttpServer;
use Ratchet\WebSocket\WsServer;
use MyApp\MyAppClassChat;

require dirname(__DIR__) . '/vendor/autoload.php';

$server = IoServer::factory(new HttpServer(new WsServer(new MyAppClass())), 8080);
$server->run();
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MyAppClass非常简单:

 <?php
namespace MyAppClass;
use Ratchet\MessageComponentInterface;
use Ratchet\ConnectionInterface;

class MyAppClass implements MessageComponentInterface {

    protected $clients;

    public function __construct() {
        $this->clients = new \SplObjectStorage;
    }

    public function onOpen(ConnectionInterface $conn) {
            /* I would like to put recover the session infos of the clients here
               but the session_start() call returns an empty array ($_SESSION variables have been previuosly set by the web server)*/
        session_start();
        var_dump($_SESSION) // empty array...
        echo "New connection! ({$conn->resourceId})\n";
    }

    public function onMessage(ConnectionInterface $from, $msg) {
        $numberOfReceivers = count($this->clients) -1;
        echo sprintf('Connection %d sending message "%s" to %d other connection%s' . "\n", $from->resourceId, $msg, 
                                 $numberOfReceivers, $numberOfReceivers == 1 ? '' : 's');

        $this->clients->rewind();
        while ($this->clients->valid())
        {
            $client = $this->clients->current();
            if ($client !== $from) {
                $client->send($msg);
            }
            $this->clients->next();
        }
    }

    public function onClose(ConnectionInterface $conn) {
        $this->clients->detach($conn);
        echo "Connection {$conn->resourceId} has disconnected\n";
    }

    public function onError(ConnectionInterface $conn, \Exception $e) {
        echo "An error has occurred: {$e->getMessage()}\n";
        $conn->close();
    }
}
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有没有办法用我的实际布局做到这一点,还是应该配置apache才能使用mod_proxy_wstunnel模块?

感谢帮助!!!

dmi*_*309 12

正如其他StackOverflow答案所示(Ratchet没有Symfony会话,在棘轮websocket连接中启动会话),没有办法直接在Apache和Ratchet进程之间共享$ _SESSION变量.但是,可以启动与Apache服务器的会话,然后访问Ratchet代码中的会话cookie.

Apache服务器的index.html启动会话:

<?php
// Get the session ID.
$ses_id = session_id();
if (empty($ses_id)) {
    session_start();
    $ses_id = session_id();
}
?><!DOCTYPE html> ...
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Ratchet MessageComponentInterface代码访问会话令牌:

public function onMessage(ConnectionInterface $from, $msg) {
    $sessionId = $from->WebSocket->request->getCookies()['PHPSESSID'];
    # Do stuff with the token...
}
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一旦两个服务器都知道用户的会话令牌,他们就可以使用令牌通过MySQL数据库共享信息(这就是我所做的):

    # Access session data from a database:
    $stmt = $this->mysqli->prepare("SELECT * FROM users WHERE cookie=?");
    $stmt->bind_param('s', $sessionId);
    $stmt->execute();
    $result = $stmt->get_result();
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或者,您可以进行更奇特的进程间通信:

    # Ratchet server:
    $opts = array(
        'http'=>array(
            'method'=>'GET',
            'header'=>"Cookie: PHPSESSID=$sessionId\r\n"
        )
    );
    $context = stream_context_create($opts);
    $json = file_get_contents('http://localhost:80/get_session_info.php', false, $context);
    $session_data = json_decode($json);

    # Apache server's get_session_info.php
    # Note: restrict access to this path so that remote users can't dump
    # their own session data.
    echo json_encode($_SESSION);
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