disk_free_space
Something that might go well with this function is the ability to list available disks. On Windows, here's the relevant code:
<?php
/**
* Finds a list of disk drives on the server.
* @return array The array velues are the existing disks.
*/
function get_disks(){
if(php_uname('s')=='Windows NT'){
// windows
$disks=`fsutil fsinfo drives`;
$disks=str_word_count($disks,1);
if($disks[0]!='Drives')return '';
unset($disks[0]);
foreach($disks as $key=>$disk)$disks[$key]=$disk.':\\';
return $disks;
}else{
// unix
$data=`mount`;
$data=explode(' ',$data);
$disks=array();
foreach($data as $token)if(substr($token,0,5)=='/dev/')$disks[]=$token;
return $disks;
}
}
?>
EXAMPLE OF USE:
<?php print_r(get_disks()); ?>
EXAMPLE RESULT:
Array
(
[1] => A:\
[2] => C:\
[3] => D:\
[4] => E:\
[5] => F:\
[6] => G:\
[7] => H:\
[8] => I:\
[9] => M:\
[10] => X:\
[11] => Z:\
)
Warning: This also finds empty disk drives (eg; CD or SMD drives or the more common floppy drive).
Warning2: If you want to find space usage using the info from my function, prepend the disk function with the "@", eg:
$free=@disk_free_space('A:\\');
disk_total_space
(PHP 4 >= 4.1.0, PHP 5)
disk_total_space — Retourne la taille d'un dossier
Description
float disk_total_space
( string $directory
)
Lit récursivement toutes les tailles du dossier directory et retourne la somme en octets.
Liste de paramètres
- directory
-
Un dossier du système de fichier ou la partition d'un disque.
Valeurs de retour
Retourne la taille en octets, sous la forme d'un nombre décimal ou FALSE si une erreur survient.
Exemples
Exemple #1 Exemple avec disk_total_space()
<?php
// $df contient le nombre d'octets du dossier "/"
$df = disk_total_space("/");
// Sous Windows :
disk_total_space("C:");
disk_total_space("D:");
?>
Notes
Note: Cette fonction ne fonctionne pas avec les fichiers distants, car le fichier utilisé doit être accessible sur le système de fichiers local.
add a note
User Contributed Notesdisk_total_space
cotact[at]covac-software[dot]com
25-Dec-2009 03:52
25-Dec-2009 03:52
Viitala
04-Feb-2008 10:04
04-Feb-2008 10:04
Beware of empty files!
<?php
// Wrong
$exp = floor(log($bytes) / log(1024));
//Correct
$exp = $bytes ? floor(log($bytes) / log(1024)) : 0;
?>
tularis at php dot net
25-Jun-2007 12:13
25-Jun-2007 12:13
For a non-looping way to add symbols to a number of bytes:
<?php
function getSymbolByQuantity($bytes) {
$symbols = array('B', 'KiB', 'MiB', 'GiB', 'TiB', 'PiB', 'EiB', 'ZiB', 'YiB');
$exp = floor(log($bytes)/log(1024));
return sprintf('%.2f '.$symbol[$exp], ($bytes/pow(1024, floor($exp))));
}
stierguy1 at msn dot com
25-Jun-2007 04:03
25-Jun-2007 04:03
function roundsize($size){
$i=0;
$iec = array("B", "Kb", "Mb", "Gb", "Tb");
while (($size/1024)>1) {
$size=$size/1024;
$i++;}
return(round($size,1)." ".$iec[$i]);}
nikolayku at tut dot by
17-Apr-2007 10:16
17-Apr-2007 10:16
Very simple function that convert bytes to kilobytes, megabytes ...
function ConvertBytes($number)
{
$len = strlen($number);
if($len < 4)
{
return sprintf("%d b", $number);
}
if($len >= 4 && $len <=6)
{
return sprintf("%0.2f Kb", $number/1024);
}
if($len >= 7 && $len <=9)
{
return sprintf("%0.2f Mb", $number/1024/1024);
}
return sprintf("%0.2f Gb", $number/1024/1024/1024);
}
Mat
05-Apr-2007 12:46
05-Apr-2007 12:46
JulieC:
I think you may have misunderstood - given a directory, this function tells you how big the disk paritition is that the directory exists on.
So disk_total_space("C:\Windows\") will tell you how big drive C is.
It is not suggesting that a directory is a disk partition.
JulieC
31-Jan-2007 04:11
31-Jan-2007 04:11
"filesystem or disk partition" does not equal "directory" for Windows. Thanks.
martijn at mo dot com
13-Jan-2007 11:41
13-Jan-2007 11:41
This works for me (on a UNIX server):
<?php
function du( $dir )
{
$res = `du -sk $dir`; // Unix command
preg_match( '/\d+/', $res, $KB ); // Parse result
$MB = round( $KB[0] / 1024, 1 ); // From kilobytes to megabytes
return $MB;
}
$dirSize = du('/path/to/dir/');
?>
shalless at rubix dot net dot au
16-Jul-2003 04:36
16-Jul-2003 04:36
My first contribution. Trouble is the sum of the byte sizes of the files in your directories is not equal to the amount of disk space consumed, as andudi points out. A 1-byte file occupies 4096 bytes of disk space if the block size is 4096. Couldn't understand why andudi did $s["blksize"]*$s["blocks"]/8. Could only be because $s["blocks"] counts the number of 512-byte disk blocks not the number of $s["blksize"] blocks, so it may as well just be $s["blocks"]*512. Furthermore none of the dirspace suggestions allow for the fact that directories are also files and that they also consume disk space. The following code dskspace addresses all these issues and can also be used to return the disk space consumed by a single non-directory file. It will return much larger numbers than you would have been seeing with any of the other suggestions but I think they are much more realistic:
<?php
function dskspace($dir)
{
$s = stat($dir);
$space = $s["blocks"]*512;
if (is_dir($dir))
{
$dh = opendir($dir);
while (($file = readdir($dh)) !== false)
if ($file != "." and $file != "..")
$space += dskspace($dir."/".$file);
closedir($dh);
}
return $space;
}
?>
andudi at gmx dot ch
12-Jun-2002 12:15
12-Jun-2002 12:15
To find the total size of a file/directory you have to differ two situations:
(on Linux/Unix based systems only!?)
you are interested:
1) in the total size of the files in the dir/subdirs
2) what place on the disk your dir/subdirs/files uses
- 1) and 2) normaly differs, depending on the size of the inodes
- mostly 2) is greater than 1) (in the order of any kB)
- filesize($file) gives 1)
- "du -ab $file" gives 2)
so you have to choose your situation!
on my server I have no rights to use "exec du" in the case of 2), so I use:
$s = stat($file);
$size = $s[11]*$s[12]/8);
whitch is counting the inodes [12] times the size of them in Bits [11]
hopes this helps to count the used disk place in a right way... :-)
Andreas Dick