asked on

List all subdirectories, subdirectory's directories as list

I have a directory structure like this one

main/category/name

each name directory has a bunch of subdirectories and files that we don't need to consider

I need a php script that can read the contents of the main folder and print out a list in a similar fashion:

<li>category
<ul>
<li>name1</li>
<li>name2></li>
<li>namen></li>
</ul>
</li>
<li>category2
<ul>
<li>name1</li>
<li>name2></li>
<li>namen></li>
</ul>
</li>

Until no more categories are found. How can i achieve this?

Thank you very much

MMDeveloper

what version of PHP?

MMDeveloper

ok well either way, try this script, just feed it the directory path..

<?php
print_r(makeULLI(readDirR("./skins")));
 
function readDirR($dir = "./") {
	$listing = opendir($dir);
	$return = array ();
	while(($entry = readdir($listing)) !== false) {
		if ($entry != "." && $entry != "..") {
			$item = $dir . "/" . $entry;
			//echo $item . "<br />";
			if (is_file($item)) {
				$return[] = $entry;
			}
			elseif (is_dir($item)) {
				$return[$entry] = readDirR($item);
			}
		} else {}
	}
 
	return $return;
}
 
function makeULLI($array) {
	$return = "<ul>";
 
	if (is_array($array) && count($array) > 0) {
		foreach ($array as $k => $v) {
			if (is_array($v)) {
				$return .= "<li>" . $v . makeULLI($v) . "</li>";
			}
			else {
				$return .= "<li>" . $v . "</li>";
			}
		}
	} else {}
 
	$return .= "</ul>";
 
	return $return;
}
?>

Open in new window

MMDeveloper

sorry you didn't want the files to be there. here is a modified version that only looks at directories.

<?php
print_r(makeULLI(readDirR("./skins")));
 
function readDirR($dir = "./") {
	$listing = opendir($dir);
	$return = array ();
	while(($entry = readdir($listing)) !== false) {
		if ($entry != "." && $entry != "..") {
			$item = $dir . "/" . $entry;
			if (is_file($item)) {
				//$return[] = $entry;
			}
			elseif (is_dir($item)) {
				$return[$entry] = readDirR($item);
			}
		} else {}
	}
 
	return $return;
}
 
function makeULLI($array) {
	$return = "<ol>";
 
	if (is_array($array) && count($array) > 0) {
		foreach ($array as $k => $v) {
			if (is_array($v)) {
				$return .= "<li>" . $k . makeULLI($v) . "</li>";
			}
			else {
				$return .= "<li>" . $v . "</li>";
			}
		}
	} else {}
 
	$return .= "</ol>";
 
	return $return;
}
?>

Open in new window

0plus1

ASKER

php version: 5

Your script fails because prints this:

# Array

# Array

# Array

* Array
o nomefile
o nomefile
o nomefile

# Array

# Array

A parte fromt the array error, it shouldn't print the file inside the subfolder, instead it should ignore them completely

MMDeveloper

try the second post, that problem was fixed in it :P

0plus1

ASKER

We are getting there, the problem now is this i want exactly this fashion:

<li>category
<ul>
<li>name1</li>
<li>name2></li>
<li>namen></li>
</ul>
</li>

while your script outputs this:

<ul>
<li>category
<ul>
<li>name1<ul></ul></li>
<li>name2<ul></ul></li>
</ul>
</li>

ASKER CERTIFIED SOLUTION

MMDeveloper

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

0plus1

ASKER

Thank you man
almost there (sorry to be so finicky) the last thing is that i don't want the first <ul> tag, i know it seems strange and it's not valid html but this is going to be part of a suckefish style menu and i need a custum ul tag at the beginning. So to get ONLY the first <ul> tag what should i do? Put an if condition?

Thanks again man, you are my saviour..

SOLUTION

MMDeveloper

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

0plus1

ASKER

Thank you again, you spared me a lot of time.. i was stuck on this one!