PHP - Can't Open A Folder

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http://www.phpfreaks.com/forums/index.php?topic=308625.0

Hello all, i am go9090go.
Today i made a domains for a jar file people can upload from my website.
I made this to make the jar file close source and its easy to update.
Now i made a java classloader and everything i made works.
The classloader call a php document with the password and username.
The pass and name will be checked inside a databse and if its inside i use
header() to load the jar file.

But when i just go to my main domain i get the index of the site and people can easly download the jar file without have to walk thru the php pass checker.
So i want to place the jar file inside a protected folder,and i want that only way you get acces to this jar is by the php file. How can i get a file from a protected folder?

here is the php used when the jar file is not inside a protected folder:

<?php

$DBName = "name";//name database        
$DBUser = "name";//user        
$DBPassword = "pass"; //passs
$DBHost = "host"; //might be different
         
mysql_connect($DBHost, $DBUser, $DBPassword);
        mysql_select_db($DBName);

        $username = $_GET['username'];
        $password = $_GET['password'];

$IP = $_SERVER['REMOTE_ADDR'];
    
$string = "Java";
$pos = strpos($agent, $string);    
if (!strpos($_SERVER['HTTP_USER_AGENT'], "Java")) 
{        
echo("Your Auth has been banned for trying to breach security.");        
//mysql_query("delete from users where username='$username'");
exit();    
}

$query = "select * from users where name='$username' and pass='$password'";
        mysql_query($query);
        $num = mysql_affected_rows();
if ($num > 0) 
{
header('Location:script/Script.jar');
}

?>


now i want to use the header to a file inside a folder that is protected :



so how can i make the header() methode to open script.jar inside a protected folder.
The folder haves name and pass: blabla,balbla for exempel

thanks for help

Hi,   I'm newbie. My requirement is to open network shared folder from PHP. This works fine in IE but not in Chrome. Can anyone please help me ....   Thanks

I got this script:

But it give me error, file_get_contents cannot open stream.

I need to add the FTP connection with user/pass paramaters.

then look in set http url, to get the file contents(images) and transfer to ftp server location.

Can Anyone take alook and tell me if I am going down the right path and how to get there. Please

Code: [Select]
function postToHost($host, $port, $path, $postdata = array(), $filedata = array()) {
$data = "";
$boundary = "---------------------".substr(md5(rand(0,32000)),0,10);
$fp = fsockopen($host, $port);

fputs($fp, "POST $path HTTP/1.0\n");
fputs($fp, "Host: $host\n");
fputs($fp, "Content-type: multipart/form-data; boundary=".$boundary."\n");

// Ab dieser Stelle sammeln wir erstmal alle Daten in einem String
// Sammeln der POST Daten
foreach($postdata as $key => $val){
$data .= "--$boundary\n";
$data .= "Content-Disposition: form-data; name=\"".$key."\"\n\n".$val."\n";
}

// Sammeln der FILE Daten
if($filedata) {
$data .= "--$boundary\n";
$data .= "Content-Disposition: form-data; name=\"".$filedata['name']."\"; filename=\"".$filedata['name']."\"\n";
$data .= "Content-Type: ".$filedata['type']."\n";
$data .= "Content-Transfer-Encoding: binary\n\n";
$data .= $filedata['data']."\n";
$data .= "--$boundary--\n";
}

// Senden aller Informationen
fputs($fp, "Content-length: ".strlen($data)."\n\n");
fputs($fp, $data);

// Auslesen der Antwort
while(!feof($fp)) {
$res .= fread($fp, 1);
}
fclose($fp);

return $res;
}

$postdata = array('var1'=>'today', 'var2'=>'yesterday');
$filedata = array(
'type' => 'image/png',
'data' => file_get_contents('http://xxx/tdr-images/images/mapping/dynamic/deals/spot_map')
);

echo '<pre>'.postToHost ("localhost", 80, "/test3.php", $postdata, $filedata).'</pre>';

I'm trying to open a URL using fopen() and I keep getting this error:
Code: [Select]
Failed to open stream: Permission deniedI can not work out what is wrong, it's not a firewall as far as I can tell.