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List all subdirectories, subdirectory's directories as list

Posted on 2008-10-15
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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
  • 6
  • 4
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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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
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After years of analyzing threat actor behavior, it’s become clear that at any given time there are specific tactics, techniques, and procedures (TTPs) that are particularly prevalent. By analyzing and understanding these TTPs, you can dramatically enhance your security program.

 

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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0
 

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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