disk_free_space
Beware of empty files!
<?php
// Wrong
$exp = floor(log($bytes) / log(1024));
//Correct
$exp = $bytes ? floor(log($bytes) / log(1024)) : 0;
?>
disk_total_space
(PHP 4 >= 4.1.0, PHP 5)
disk_total_space — ディレクトリの全体サイズを返す
説明
float disk_total_space
( string $directory
)
ディレクトリを含む文字列を指定してください。この関数は、 ファイルシステムまたはディスクパーティションに対応する全体バイト数を返します。
パラメータ
- directory
-
ファイルシステムのディレクトリあるいはディスクパーティション。
返り値
総バイト数を float 型で返します。
例
例1 disk_total_space() の例
<?php
// $df は、「/」で利用可能な全体バイト数
$df = disk_total_space("/");
// Windows の場合:
disk_total_space("C:");
disk_total_space("D:");
?>
注意
注意: この関数では、 リモートファイル を 使用することはできません。これは、処理されるファイルがサーバの ファイルシステムによりアクセスできる必要があるためです。
add a note
User Contributed Notesdisk_total_space
Viitala
04-Feb-2008 09:04
04-Feb-2008 09:04
tularis at php dot net
25-Jun-2007 10:13
25-Jun-2007 10:13
For a non-looping way to add symbols to a number of bytes:
<?php
function getSymbolByQuantity($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 02:03
25-Jun-2007 02:03
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 08:16
17-Apr-2007 08:16
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 10:46
05-Apr-2007 10:46
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 03:11
31-Jan-2007 03:11
"filesystem or disk partition" does not equal "directory" for Windows. Thanks.
martijn at mo dot com
13-Jan-2007 10:41
13-Jan-2007 10:41
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 02:36
16-Jul-2003 02:36
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 $space;
}
?>
andudi at gmx dot ch
11-Jun-2002 10:15
11-Jun-2002 10:15
To find the total size of a file/directory you have to differ two situations:
(on Linux/Unix based systems only!?)
you are interested:
1) in the total size of the files in the dir/subdirs
2) what place on the disk your dir/subdirs/files uses
- 1) and 2) normaly differs, depending on the size of the inodes
- mostly 2) is greater than 1) (in the order of any kB)
- filesize($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