Something like so:

/* sendToHost
 * ~~~~~~~~~~
 * Params:
 *   $proxy     - Proxy you want to use
 *   $host      - Just the hostname.  No http:// or
                  /path/to/file.html portions
 *   $method    - get or post, case-insensitive
 *   $path      - The /path/to/file.html part
 *   $data      - The query string, without initial question mark
 *   $useragent - If true, 'MSIE' will be sent as
                  the User-Agent (optional)
 *
 * Examples:
 *   sendToHost('webcache', 'www.google.com','get','/search','q=php_imlib');
 *   sendToHost('localhost', 'www.example.com','post','/some_script.cgi',
 *              'param=First+Param&second=Second+param');
 */

function sendToHost($proxy, $host,$method,$path,$data,$useragent=0)
{
    // Supply a default method of GET if the one passed was empty
    if (empty($method)) {
        $method = 'GET';
    }
    $method = strtoupper($method);
    $fp = fsockopen($proxy, 8080);
    if ($method == 'GET') {
      $path .= '?' . $data;
    }
    fputs($fp, "$method http://$host/$path HTTP/1.0\r\n");
    fputs($fp, "Host: $host\r\n");
    if ($useragent) {
      fputs($fp, "User-Agent: MSIE\r\n");
    }
    if ($method == 'POST') {
      fputs($fp,"Content-type: application/x-www-form-urlencoded\r\n");
      fputs($fp, "Content-length: " . strlen($data) . "\r\n");
    }
    fputs($fp, "Connection: close\r\n\r\n");
    if ($method == 'POST') {
        fputs($fp, $data);
    }

    while (!feof($fp)) {
        $buf .= fgets($fp,128);
    }
    fclose($fp);
    return $buf;
}

I've not tried this code, but I think I got it right :)
Also please note that this connects to proxys on port 8080, you might need
to change that.
Oh and I changed the http/1.1 to 1.0 because I find it has less problems :)

Hope this works/helps
Andrew
----- Original Message -----
From: "David Yee" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, July 28, 2003 3:44 AM
Subject: [PHP] POST/GET using a proxy server


> Hi all- how do I modify the following function to work over a proxy
server?
> Thanks for any help.
>
> David


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