Finding Local Disk Space

[navjak]

Hi Guys,

 

Can any one tell me how to find the local disk space in php?

I tried the command df -h, it gives me the details of the space for local drives as well as for all the mapped drives. All I want is to get the disk space of local drives. df -l, doesnot work. I can get the space details if I hardcode the drive letter.

Need to find the local disk space without hardcoding the drive letters.

 

Please help me......

 

Thanks & Regards,

navjak.

[redarrow]

<?php

$x=disk_total_space("C:");

echo $x;
?> 

 

link

http://uk.php.net/manual/en/function.disk-total-space.php

[navjak]

Thanks for the responce.

 

-----Need to find the local disk space without hardcoding the drive letters.

 

As I mentioned, I can get the diskspace of the local drive if I hardcode it. Need to find out the disk space of the local drives without hardcoding the drive letters.

 

navjak

[redarrow]

ok i tried everythink to solve your problam

i found that there meny errors using the disk_total_space function

so the only way was to do it as below ok.

 

dont forget the error_reporting will have to be turned off on this code ok.

 

 

this code will show on any windows meachine the drives and size of all drives a-z

not renamed drives.

<?php
error_reporting(0);

$a=disk_total_space("a:");
$b=disk_total_space("b:");
$c=disk_total_space("c:");
$d=disk_total_space("d:");
$e=disk_total_space("e:");
$f=disk_total_space("f:");
$g=disk_total_space("g:");
$h=disk_total_space("h:");
$i=disk_total_space("i:");
$j=disk_total_space("j:");
$k=disk_total_space("k:");
$l=disk_total_space("l:");
$m=disk_total_space("m:");
$n=disk_total_space("n:");
$o=disk_total_space("o:");
$p=disk_total_space("p:");
$q=disk_total_space("q:");
$r=disk_total_space("r:");
$s=disk_total_space("s:");
$t=disk_total_space("t:");
$u=disk_total_space("u:");
$v=disk_total_space("v:");
$w=disk_total_space("w:");
$x=disk_total_space("x:");
$y=disk_total_space("y:");
$z=disk_total_space("z:");

if($a){$a="disk a: ".disk_total_space("a:")." ";}

if($b){$b="disk b: ".disk_total_space("b:")." ";}

if($c){$c="disk c: ".disk_total_space("c:")." ";}

if($d){$d="disk d: ".disk_total_space("d:")." ";}

if($e){$e="disk e: ".disk_total_space("e:")." ";}

if($f){$f="disk f: ".disk_total_space("f:")." ";}

if($g){$g="disk g: ".disk_total_space("g:")." ";}

if($h){$h="disk h: ".disk_total_space("h:")." ";}

if($i){$i="disk i: ".disk_total_space("i:")." ";}

if($j){$j="disk j: ".disk_total_space("j:")." ";}

if($k){$k="disk k: ".disk_total_space("k:")." ";}

if($l){$l="disk l: ".disk_total_space("l:")." ";}

if($m){$m="disk m: ".disk_total_space("m:")." ";}

if($n){$n="disk n: ".disk_total_space("n:")." ";}

if($o){$o="disk o: ".disk_total_space("o:")." ";}

if($p){$p="disk p: ".disk_total_space("p:")." ";}

if($q){$q="disk q: ".disk_total_space("q:")." ";}

if($r){$r="disk r: ".disk_total_space("r:")." ";}

if($s){$s="disk s: ".disk_total_space("s:")." ";}

if($t){$t="disk t: ".disk_total_space("t:")." ";}

if($u){$u="disk u: ".disk_total_space("u:")." ";}

if($v){$v="disk v: ".disk_total_space("v:")." ";}

if($w){$w="disk w: ".disk_total_space("w:")." ";}

if($x){$x="disk x: ".disk_total_space("x:")." ";}

if($y){$y="disk y: ".disk_total_space("y:")." ";}

if($z){$z="disk z: ".disk_total_space("z:")." ";}

echo"$a <br>$b<br>$c<br>$d<br>$e<br>$f<br>$g<br>$h<br>
$i<br>$j<br>$k<br>$l<br>$m <br>$n<br>$o<br>$p<br>$q<br>
$u<br>$v<br>$w<br>$x<br>$y<br>$z<br>";
?> 

[DeathStar]

I would rather make it check with a foreach loop.

Much less scripting and much faster.

[redarrow]

you did not read my post wont work with any loop that function dont you think i would off joke lol.

[DeathStar]

Oh, okay.

Sorry.

 

you did not read my post wont work with any loop that function dont you think i would off joke lol.

[navjak]

sorry guys......I think Iam not clear enough of wht I want. I said I dont want to hardcode.

Let me explain it clearly wht Iam doing n wht I want.

 

I had written a php script on linux machine to find the disk space on windows machines(which is remote).

I had used ssh to run the df command.

 

"ssh -l <userid> <machinename> df -h c:"

 

This gives me the size,used,available and used % of the drive C. Similarly for D

 

"ssh -l <userid> <machinename> df -h d:"

if the local disk is partitioned.

 

If I dont mention any drive letter at the end

"ssh -l <userid> <machinename> df -h "

it gives the space for all the mapped drives + local drives(which are c/d/e/f.....).

 

For some machines the local disk may be any letter, which I cant know.

So I want to know the space of the local disks ONLY. I dont want to display all the drive spaces which are mapped on the remote machine.

 

Please help me..................

 

Thanks & Regards,

navjak

[trq]

What do you mean when you say df -l is not working? What output do you get?

[navjak]

I just get the below message. Thats it. It does not display any size of the local drives.

 

Filesystem          1K-blocks      Used Available Use% Mounted on

[navjak]

guys any update on the above issue??

[navjak]

I tried reinstalling Cygwin and used df -l. It gave me the result with local drives and also with the Cygwin path directories.(bin and lib directories)

How can I remove the cygwin path directories and get only the local drive space??