Solved

List all subdirectories, subdirectory's directories as list

Posted on 2008-10-15
10
1,097 Views
Last Modified: 2008-10-15
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
0
Comment
Question by:0plus1
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10 Comments
 
LVL 15

Expert Comment

by:MMDeveloper
ID: 22721753
what version of PHP?
0
 
LVL 15

Expert Comment

by:MMDeveloper
ID: 22721953
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;
}
?>

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0
 
LVL 15

Expert Comment

by:MMDeveloper
ID: 22721979
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;
}
?>

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Author Comment

by:0plus1
ID: 22722004
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
0
 
LVL 15

Expert Comment

by:MMDeveloper
ID: 22722293
try the second post, that problem was fixed in it :P
0
 

Author Comment

by:0plus1
ID: 22722504
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>

0
 
LVL 15

Accepted Solution

by:
MMDeveloper earned 100 total points
ID: 22723548
ah ok it was trying to print a listing of an empty directory. I have modified the code and this outputs valid html (per your example) for me.
<?php
print_r(makeULLI(readDirR()));
 
function readDirR($dir = "./") {
	$listing = opendir($dir);
	$return = array ();
	while(($entry = readdir($listing)) !== false) {
		if ($entry != "." && $entry != "..") {
			$item = $dir . "/" . $entry;
			if (is_dir($item)) {
				$return[$entry] = readDirR($item);
			} else {}
		} else {}
	}
 
	return $return;
}
 
function makeULLI($array) {
	$return = "<ol>\n";
 
	if (is_array($array) && count($array) > 0) {
		foreach ($array as $k => $v) {
			if (is_array($v) && count($v) > 0) {
				$return .= "\t<li>" . $k . makeULLI($v) . "</li>\n";
			}
			else {
				$return .= "\t<li>" . $k . "</li>\n";
			}
		}
	} else {}
 
	$return .= "</ol>";
 
	return $return;
}
?>

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Author Comment

by:0plus1
ID: 22723863
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..
0
 
LVL 15

Assisted Solution

by:MMDeveloper
MMDeveloper earned 100 total points
ID: 22724286
yeah I'd try this modified function
function makeULLI($array, $printul = false) {
	if ($printul != false) {
        	$return = "<ol>\n";
	} else {}
 
        if (is_array($array) && count($array) > 0) {
                foreach ($array as $k => $v) {
                        if (is_array($v) && count($v) > 0) {
                                $return .= "\t<li>" . $k . makeULLI($v, true) . "</li>\n";
                        }
                        else {
                                $return .= "\t<li>" . $k . "</li>\n";
                        }
                }
        } else {}
 
	if ($printul != false) {
        	$return .= "</ol>";
	} else {}
 
        return $return;
}

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0
 

Author Comment

by:0plus1
ID: 22724710
Thank you again, you spared me a lot of time.. i was stuck on this one!
0

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