How To Sort Files Of A Directory using Php

itssami

May 1 2006, 11:30 AM

The following code displays the files of folder...but they are displaced by the order of adding...
i want to sord the files / folders alphabetically and sord by accending order and by decending order..
can some one help me.

CODE

<?
$path = "";
$dir_handle = @opendir($path) or die("Unable to open $path");
echo "Directory Listing of $path<br/>";
while($file = readdir($dir_handle)) {
    if(is_dir($file)) {
        continue; 
        }

   else if($file != '.' && $file != '..') {
              echo "<a href='$path/$file'>$file</a><br/>";
            }
}



//closing the directory
closedir($dir_handle);

?>

Use [code][/code] tags for code!

mole2k9

May 1 2006, 12:34 PM

haven't tested this but give it a try should work, create an array of the file names and then use the sort function.

CODE

<?
$path = "";
$dir_handle = @opendir($path) or die("Unable to open $path");
echo "Directory Listing of $path ";
$i=0;
while($file = readdir($dir_handle))
{
 if(is_dir($file))
 {
 continue;
 }
 else if($file != '.' && $file != '..')
 {
 //echo "<a href='$path/$file'>$file</a> ";
 $narray[$i]=$file;
 $i++;
 }
}
sort($narray);
for($i=0;i<sizeof($narray);$i++)
{
 echo "<a href=".chr(34).$path."\".$narray[$i].chr(34).">".$file."</a> ";
}
//closing the directory
closedir($dir_handle);
?>

electriic ink

May 1 2006, 12:37 PM

So you want to display the files/folders in a directory alphabetically? The code is as follows:

CODE

<?
$path = "";
$files_gathered = array();

$dir_handle = @opendir($path) or die("Unable to open $path");
echo "Directory Listing of $path<br/>";

 while($file = readdir($dir_handle)) {
 
  if(is_dir($file)) {
   continue;
  } else if($file != '.' && $file != '..') {
   $files_gathered[] = $file;
  }

}

//closing the directory
closedir($dir_handle);

// begin sorting and displaying the files

 $files_gathered = sort ($files_gathered);
 $count = count ($files_gathered);

 for ($loop_start = 0; isset($files_gathered[$loop_start]); $loopstart++) {
  echo "<a href='" . $path . $files_gathered[$loop_start] . 
"' title="' . $files_gathered[$loop_start] . "'> " . $files_gathered[$loop_start] . " </a> <br />";
 } 
 
?>

I haven't tested it yet but it should still work

jlhaslip

May 1 2006, 01:26 PM

Both of those scripts look like they will work to sort in ascending order.
To sort the array in descending order, use the function rsort() instead of sort(). Everything else should remain the same.

mole2k9

May 1 2006, 03:10 PM

Opps mean to use rsort also made 1 other change , chnage$sfile to $narray[$i] in,

echo "<a href=".chr(34).$path.$narray[$i].chr(34).">".$narray[$i]."</a> ";

this is tested and works,

CODE

<?
$path = "";
$narray=array();
$dir_handle = @opendir($path) or die("Unable to open $path");
echo "Directory Listing of $path ";
$i=0;
while($file = readdir($dir_handle))
{
 if(is_dir($file))
 {
 continue;
 }
 else if($file != '.' && $file != '..')
 {
 //echo "<a href='$path/$file'>$file</a> ";
 $narray[$i]=$file;
 $i++;
 }
}
rsort($narray);

for($i=0; $i<sizeof($narray); $i++)
{
echo "<a href=".chr(34).$path.$narray[$i].chr(34).">".$narray[$i]."</a> ";

}

//closing the directory
closedir($dir_handle);
?>

you might want to add a function to check whether the path has a \ at the end and add one if not.

jlhaslip

May 1 2006, 03:11 PM

All it needs now is a form to select the directory and a radio button set to select ascending or descending sort order. Then an If statement to sort in the chosen order. And a set of html ul tags and li tags. And a class to hook the styling to for the html / css and you are all set.

I prefer a single page script / form method personally.

Looks good.

itssami

May 1 2006, 03:31 PM

i dont mean to bother but it still shows blank screen... for example i have a folder "test" in htdocs.. and i gave the path

CODE

$path = "test";

but it shows blank screen.. and if i give any wrong name of the folder which doesnt even exists in htdocs , it still shows blank page, even it should say "Unable to open ..."
im trying to find the problem but cant..

QUOTE(mole2k9 @ May 1 2006, 03:10 PM)

Opps mean to use rsort also made 1 other change , chnage$sfile to $narray[$i] in,
echo "<a href=".chr(34).$path.$narray[$i].chr(34).">".$narray[$i]."</a> ";
this is tested and works,

CODE

you might want to add a function to check whether the path has a \ at the end and add one if not.

jlhaslip

May 1 2006, 03:49 PM

Have a look at the script I sent to you and you will notice there is a line in in which sets the path to "./". This references the path which the script is run from. Modify this path to include the name of the directory on a relative basis.

CODE

$path = "./test/"

Also, if you are making changes to the script and want further advise, I would reccomend including a copy of the actual code which is not working. The sample you list above does not have any directory in it so I am curious about whether you actually have a valid path in the source code.

Also, the echo statement is commented out in the listing you display.

CODE

//echo "<a href='$path/$file'>$file</a> ";

Check to make sure the commenting slashes are removed before running the script. With these slashes in place, the information is not being sent to the Browser. Might explain the blank page.

mole2k9

May 1 2006, 04:14 PM

try $path = "/test";

itssami

May 1 2006, 08:31 PM

thanks..i have edited a lil bit the above code..now it displays all the files and folders within a directory...but the problem is that , for example i have index.php file in that directory , it shows that index.php file also in the list..i want that it should show every file but NOT index.php.. what i should do for that ??

CODE

<?php
$path = "./";
$narray=array();
$dir_handle = @opendir($path) or die("Unable to open $path");
echo "Directory Listing of $path ";
$i=0;
while($file = readdir($dir_handle))
{
 if($file != '.' && $file != '..')
 {
 //echo "<a href='$path/$file'>$file</a> ";
 $narray[$i]=$file;
 $i++;
 }
}
sort($narray);

for($i=0; $i<sizeof($narray); $i++)
{
echo "<a href=".chr(34).$path.$narray[$i].chr(34).">".$narray[$i]."</a> ";

}

//closing the directory
closedir($dir_handle);
?>

Latest Entries

BuffaloHELP

Aug 25 2006, 04:30 AM

By the way you can write as

HTML

echo "<a href='$path$narray[$i]'>$narray[$i]</a> ";

and still be sufficient. As long as $path is correct you do not need char(34) since a single quote will suffice.

And when only one set of double quotes are used, no need to use periods to separate variables.

But the question I have is: how can I be able to hide the file extension? Is there a way or is it just too much?

---UPDATE---

Figured out how to replace

CODE

$filename = str_replace(".htm", ".", $narray[$i]);

CODE

$filename = str_replace(".htm", "", $narray[$i]);

Place this right before echo (in my case) solved my issue. The problem was that the replacing string had to have a value. I left it as just " " and it wasn't working.

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