help with this code please

wkdw1ll1ams · August 25, 2011

im making a film site for my "personal use" and i have around 110 films to host on my server but i do not want to rite out the film names and hyperlink them to the AVI file because that would take all day. so i tried using this:

http://pastebin.com/p8TiEfz5

this will allow me to go straight to the name of the AVI file i want to see. but i dont no where to put :shrug: im new to php.

could you put that code in here:

http://www.heypasteit.com/clip/0060

WebStyles · August 25, 2011

you need something just a little more complex than that.

That script you provided reads file names based on a given directory, but it seems you have a movie directory, and then each movie is contained within it's own folder.

you need to test each folder with is_dir or something.

WebStyles · August 25, 2011

try this (untested, may contain a few typos)

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<html>
<head>
<meta content="text/html; charset=iso-8859-1" http-equiv="Content-Type" />
<meta name="author" content="FilmZilla">
<meta name="publisher" content="FilmZilla">
<meta name="copyright" content="FilmZilla">
<style type="text/css">
@import url(styles.css);
</style>
<title>FilmZilla</title>
</head>

<body>
<div id="header">FilmZilla</div>
<div id="bar_nav">
<ul>
<a href="#">Home</a>
<a href="Donate.html">Donate</a>
</ul>
</div>
<div id="content_left">
<h1>Newest Films</h1>
Planet 51
<BR>
Scream 4
<BR>
Xmen First Class
</div>
<!--
this is where the film links are ##################################################
-->
<div id="content"><h1>Films</h1>
<?php
$basefolder = opendir('/films/');
while (($folder = readdir( $basefolder)) !== false){
if (is_dir($folder)){
	$subfolder = opendir('/films/'.$folder);
	while (($file = readdir( $subfolder)) !== false){
		echo $folder . '/'. $file.'<br>';
		echo '<a href="/films/'.$folder.'/'.$file.'">'.str_replace(".avi","",$file).'</a><br />';
	}
	closedir( $subfolder );
}
}
closedir( $basefolder );
?>
<!--this is where the film links end ##################################################
-->
</div>
<div id="bar_bottom2"><marquee>Films are all DVDRIP.  Please donate. Do not take credit for these movies!</marquee></div>

<div id="bar_bottom">© 2011   |   Copyright © FileZilla</a></div>

<!--[if IE]></div><![endif]--> 

<div style="text-align: center; font-size: 0.75em;">CopyrightÂ© FilmZilla</a>.</div></body>
<link rel="shortcut icon" href="icon.ico">
<link rel="icon" type="image/gif" href="icon.png">
</html>

silkfire · August 25, 2011

You're hosting copyrighted movies. Neat.

wkdw1ll1ams · August 26, 2011

i get this all the way down the page:

Warning: opendir(/films/,/films/) [function.opendir]: The system cannot find the file specified. (code: 2) in C:\xampp\htdocs\test.php on line 35

Warning: opendir(/films/) [function.opendir]: failed to open dir: No such file or directory in C:\xampp\htdocs\test.php on line 35

Warning: readdir() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\test.php on line 36

jcbones · August 26, 2011

I found an old function in my library that was easily retrofitted for this application. I don't really like echo'ing out of functions, but it was the easiest way to solve the unknown depths the resulting multi-dem array.

<?php
function dirContents($dir) {
if (is_dir($dir)) {	
	if ($dh = opendir($dir)) {			
		while (($file = readdir($dh)) !== false) {
			if(in_array($file,array('.','..'))) { continue; }
			if(is_dir($dir.'/'.$file)) {					
				$contents[$dir.'/'.$file] = dirContents($dir.'/'.$file);
			} else {
				$contents[] = $dir . '/' . $file;
			}
		}			
		closedir($dh);
	}
}
return $contents;
}

function processArray($array) {
if(is_array($array)) {
	echo '<ol>';
	foreach($array as $key => $value) {
		if(is_int($key)) {
			echo '<li><a href="'.$value.'">'.str_replace('/','',strstr(strrchr($value,'/'),'.',true)).'</a></li>';
		}
		else {
			echo '<li><span style="font-weight:bold">' . $key . '</span>';
			processArray($value);
			echo '</li>';
		}
	}
	echo '</ol>';
}
}
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<html>
<head>
<meta content="text/html; charset=iso-8859-1" http-equiv="Content-Type" />
<meta name="author" content="FilmZilla">
<meta name="publisher" content="FilmZilla">
<meta name="copyright" content="FilmZilla">
<style type="text/css">
@import url(styles.css);
</style>
<title>FilmZilla</title>
</head>

<body>
<div id="header">FilmZilla</div>
<div id="bar_nav">
<ul>
<a href="#">Home</a>
<a href="Donate.html">Donate</a>
</ul>
</div>
<div id="content_left">
<h1>Newest Films</h1>
Planet 51
<BR>
Scream 4
<BR>
Xmen First Class
</div>
<!--
this is where the film links are ##################################################
-->
<div id="content"><h1>Films</h1>
<?php
$dir = 'test';
$files = dirContents($dir);
processArray($files);
?>
<!--this is where the film links end ##################################################
-->
</div>
<div id="bar_bottom2"><marquee>Films are all DVDRIP.  Please donate. Do not take credit for these movies!</marquee></div>

<div id="bar_bottom">© 2011   |   Copyright © FileZilla</a></div>

<!--[if IE]></div><![endif]--> 

<div style="text-align: center; font-size: 0.75em;">CopyrightÂ© FilmZilla</a>.</div></body>
<link rel="shortcut icon" href="icon.ico">
<link rel="icon" type="image/gif" href="icon.png">
</html>

wkdw1ll1ams · August 26, 2011

ok, im totally new to php and i hardly know how it works. do i paste that php in a new file and save it as "download.php"

<a href="www.server.com/download.php">Scream 4</a>

and the AVI file must be the same name as the hyperlink?

jcbones · August 26, 2011

In the above script I gave you, rename the $dir variable to the folder you want. ie. films. Sorry, I left it in test mode.

$dir = 'path/to/folder'; //without trailing slash.

The link will be the same name as the avi filename, without the extension, and will link directly to that file. Try it out, if you have questions on how it works, or the formatting, post the specific question.

wkdw1ll1ams · August 26, 2011

i will give this a try. ill edit this post if it works. ty.

Edit:

Notice: Undefined variable: contents in C:\xampp\htdocs\test.php on line 16

and line 16 is

return $contents;

jcbones · August 26, 2011

Yep, that happens when you try to feed it a relative file path, with a slash BEFORE the first directory.

$dir '/films/'; //incorrect.
$dir './films'; //correct.
$dir 'films'; //correct.
$dir 'films/'; //incorrect.

wkdw1ll1ams · August 26, 2011

i removed the "/" at the end of films and still get

Notice: Undefined variable: contents in C:\xampp\htdocs\test.php on line 16

WebStyles · August 26, 2011

Notice: Undefined variable: contents

means exactly what it says: It's a Notice, not an error, and it's saying that the variable contents ($contents) is undefined. Check where $contents is used and initialize it first to avoid this error.

Apart from that, is the script working?

wkdw1ll1ams · August 26, 2011

the contents:

$contents[] = $dir . '/films' . $file;

$contents[$dir.'/films'.$file] = dirContents($dir.'/films'.$file);

if(is_dir($dir.'/films'.$file)) {

thats what i got.

WebStyles · August 26, 2011

I know what you have, I saw your code before.

$contents[] is an array that was not initialized.

wkdw1ll1ams · August 26, 2011

so how do i initialized it? :confused:

WebStyles · August 26, 2011

$contents = array();

jcbones · August 26, 2011

Ummm, I wrote the script, and I ALSO gave you the cure. If you read, it will come.

Yep, that happens when you try to feed it a relative file path, with a slash BEFORE the first directory.
$dir '/films/'; //incorrect.
$dir './films'; //correct.
$dir 'films'; //correct.
$dir 'films/'; //incorrect.

wkdw1ll1ams · August 26, 2011

i didnt understand what you mean because i tried:

$dir './films'; //correct.

$dir 'films'; //correct.

./films:

if(is_dir($dir.'/films'.$file)) {

$contents[$dir.'/films'.$file] = dirContents($dir'/films'.$file);

} else {

$contents[] = $dir . '/films' . $file;

And:

if(is_dir($dir.'films'.$file)) {

$contents[$dir.'/films'.$file] = dirContents($dir.'films'.$file);

} else {

$contents[] = $dir . 'films' . $file;

WebStyles · August 26, 2011

oh man.... :shrug:

Do you have ANY understanding of variables at all ?

WebStyles · August 27, 2011

you're trying to build a simple path to some files... you have been told that the base directory MUST be './films' or 'films'...

if you want a path like films/moviename/movie.avi and you store './films' in a variable called $dir.. then the path will be:

$dir.'/moviename/movie.avi'

wkdw1ll1ams · August 27, 2011

$contents[] = $dir . '/films' . $file;

if(is_dir($dir.'/films'.$file)) {

$contents[$dir.'/films'.$file] = dirContents($dir.'/films'.$file);

or

$contents[] = $dir . '/scream4/scream4.avi' . $file;

if(is_dir($dir.'/scream4/scream4.avi'.$file)) {

$contents[$dir.'/films'.$file] = dirContents($dir.'/scream4/scream4.avi'.$file);

like that?

WebStyles · August 27, 2011

please think about what's happening when you do this:

if(is_dir($dir.'/scream4/scream4.avi'.$file)) {

you seem very confused. Do you understand that a variable holds a piece of information, in this case a string (sequence of characters) ?

so $file holds the file's name....

and...

is_dir tests to see if the input path is a directory (folder)

now look at the code above again, and tell me what's wrong with it.

wkdw1ll1ams · August 27, 2011

hhmm. the code is wrong because the directory "films" is missing?

WebStyles · August 27, 2011

ok, back to basics... what would be the result of this?

$var = "whatever";
echo $var . ' xxx ' . $var . ' xxx';

wkdw1ll1ams · August 27, 2011

i havnt got a clue :shrug:

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