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我需要通过 php 中的 fsockopen 将会话传递给异步调用。

你能帮我把会话传递给新的套接字吗？

解决方案：

以下代码有效。

开始.php

<?php
session_start();
$_SESSION['var1'] = 'value1';
async_call('/async.php');
echo '<pre>';
print_r($_SESSION);
echo $_COOKIE['PHPSESSID'] . "\r\n";
echo '<a href="verify.php">verify.php</a>';

function async_call($filepath) {
    $host = 'sandbox'; // set to your domain
    $sock = fsockopen($host, 80);
    fwrite($sock, "GET $filepath HTTP/1.1\r\n");
    fwrite($sock, "Host: $host\r\n");
    fwrite($sock, "Cookie: PHPSESSID=" . $_COOKIE['PHPSESSID'] . "\r\n");
    fwrite($sock, "Connection: close\r\n");
    fwrite($sock, "\r\n");
    fflush($sock);
    fclose($sock);
}
?>

异步.php

<?php
session_start();
logger('confirm logger is working');
logger($_SESSION);  // this is always blank!

function logger($msg) {
    $filename = 'debug.log';
    $fd = fopen($filename, "a");
    if (is_array($msg))
        $msg = json_encode($msg);
    $msg = '[' . date('Y/m/d h:i:s', time()) . "]\t" . $msg;
    fwrite($fd, $msg . "\n");
    fclose($fd);
}
?>

验证.php

<?php
$logfile = 'debug.log';
echo '<a href="start.php">start.php</a>';
echo '<pre>' . file_get_contents($logfile);
?>

我可以让它在本地工作，但结果证明我在不允许的 fsockopen 上登台的共享主机没有任何文档。在我自己的服务器上一切都很好。谢谢您的帮助。

4

1 回答 1

1

谢谢你的帮助，安德烈亚斯。结果是我在没有任何文档的情况下在不允许的 fsockopen 上登台的共享主机。这就是为什么我可以让它在本地工作，但不能在登台服务器上工作。

以下代码有效。

开始.php

<?php
session_start();
$_SESSION['var1'] = 'value1';
async_call('/async.php');
echo '<pre>';
print_r($_SESSION);
echo $_COOKIE['PHPSESSID'] . "\r\n";
echo '<a href="verify.php">verify.php</a>';

function async_call($filepath) {
    $host = 'sandbox'; // set to your domain
    $sock = fsockopen($host, 80);
    fwrite($sock, "GET $filepath HTTP/1.1\r\n");
    fwrite($sock, "Host: $host\r\n");
    fwrite($sock, "Cookie: PHPSESSID=" . $_COOKIE['PHPSESSID'] . "\r\n");
    fwrite($sock, "Connection: close\r\n");
    fwrite($sock, "\r\n");
    fflush($sock);
    fclose($sock);
}
?>

异步.php

<?php
session_start();
logger('confirm logger is working');
logger($_SESSION);  // this is always blank!

function logger($msg) {
    $filename = 'debug.log';
    $fd = fopen($filename, "a");
    if (is_array($msg))
        $msg = json_encode($msg);
    $msg = '[' . date('Y/m/d h:i:s', time()) . "]\t" . $msg;
    fwrite($fd, $msg . "\n");
    fclose($fd);
}
?>

验证.php

<?php
$logfile = 'debug.log';
echo '<a href="start.php">start.php</a>';
echo '<pre>' . file_get_contents($logfile);
?>
于 2011-09-06T22:07:18.137 回答