PHP - Read Folder

Hi,

how to let php read the url in folder form instead of querystring?

e.g.

www.example.com/index.php?post=123
www.example.com/post/123/

both is actually directed to the same page - index.php, but how to make it read like a folder when the physical path doesn't really exist?

I am using apache web server on linux. I am using php for coding. My php code is not able to read the files from /var/tmp folder. If apache itself creates some files in /var/tmp folder then php code is able to read it. Why this permission denied issues are there though full access permissions are given to each file?

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>';

Hello, I try to get website speed of some website, but i can read only ''domain.com'' i can't read website files like css , js ... why ? i use proxies for this job. here is the php code:

$options = array(
'useragent' => "Firefox (+http://www.firefox.org)", // who am i
'connecttimeout' => 120, // timeout on connect
'timeout' => 120, // timeout on response
'redirect' => 10, // stop after 10 redirects
'referer' => "http://www.google.com",
'proxyhost' =>'85.25.8.14:80'
);
$response = http_get("http://solve-ict.com/wp-content/themes/ict%20theme/js/jquery-1.7.1.min.js", $options , $info);

but it works fine with http://domain.com/ , but with files css or js it gives 404, using some free proxy servers available ? Thanks.

Trying to get this done:

Page_1 has many external links, when certain links are clicked i do not want user to go directly to page, but rather go to a special add page_2 where user must click a second time on the link to finally get there.

The add page_2 must show on screen the name of the initial link from page_1, it must change accordingly with the link it came from page_1,once on page_2 the hyperlink redirects outside the site.

So far i am thinking give an id to the div or "<a href..." on page_1 then somehow have page_2 detect that id and fill in the variable for the final external link. Other wise is there a way to detect a url from incoming?

I guess a similar example is how some domain name sellers landing page will indicate the name of the site. Such as "Thisdomain.com" is for sale. same landing page but the name changes according to the domain that was typed.